Matematika/Iracionaliųjų funkcijų integravimas: Skirtumas tarp puslapio versijų

Dabartinė 18:13, 30 balandžio 2023 versija

Šis straipsnis yra apie iracionaliųjų funkcijų integravimą.

$\int \frac{\sqrt{x}}{x^{\frac{3}{4}} + 1} d x = \int \frac{u^{2} \cdot 4 u^{3} d u}{u^{3} + 1} = 4 [\int u^{2} d u - \int \frac{u^{2} d u}{u^{3} + 1}] = 4 [\frac{u^{3}}{3} - \frac{1}{3} \int \frac{d (u^{3} + 1)}{u^{3} + 1}] =$

$= \frac{4}{3} [u^{3} - \ln (u^{3} + 1)] + C = \frac{4}{3} [x^{\frac{3}{4}} - \ln (x^{\frac{3}{4}} + 1)] + C,$ kur $x = u^{4};$ $d x = 4 u^{3} d u .$

$\int \frac{x + x^{\frac{2}{3}} + x^{\frac{1}{6}}}{x (1 + x^{\frac{1}{3}})} d x = 6 \int \frac{(t^{6} + t^{4} + t) t^{5}}{t^{6} (1 + t^{2})} d t = 6 \int \frac{t^{5} + t^{3} + 1}{t^{2} + 1} d t = 6 \int t^{3} d t + 6 \int \frac{d t}{t^{2} + 1} =$

$= \frac{3}{2} t^{4} + 6 \arctan t + C = \frac{3}{2} x^{\frac{2}{3}} + 6 \arctan x^{\frac{1}{6}} + C,$ kur $x = t^{6};$ $d x = 6 t^{5} d t .$

$\int \frac{2}{(2 - x)^{2}} (\frac{2 - x}{2 + x})^{\frac{1}{3}} d x = - \int \frac{2 (1 + t^{3})^{2} t \cdot 12 t^{2}}{16 t^{6} (1 + t^{3})^{2}} d t = - \frac{3}{2} \int \frac{d t}{t^{3}} = \frac{3}{4 t^{2}} + C = \frac{3}{4} (\frac{2 + x}{2 - x})^{\frac{2}{3}} + C,$ kur $\frac{2 - x}{2 + x} = t^{3};$ $x = \frac{2 - t^{3}}{1 + t^{3}};$ $2 - x = \frac{4 t^{3}}{1 + t^{3}};$ $- d x = \frac{12 t^{2} (1 + t^{3}) - 4 t^{3} \cdot 3 t^{2}}{(1 + t^{3})^{2}} d t = \frac{12 t^{2} + 12 t^{5} - 12 t^{5}}{(1 + t^{3})^{2}} d t = \frac{12 t^{2}}{(1 + t^{3})^{2}} d t;$ $d x = \frac{- 12 t^{2}}{(1 + t^{3})^{2}} d t .$

$\int \frac{d x}{((x - 1)^{3} (x + 2)^{5})^{\frac{1}{4}}} = - \int \frac{(t^{4} - 1) (t^{4} - 1) 12 t^{3} d t}{3 \cdot 3 t^{4} t (t^{4} - 1)^{2}} = - \frac{4}{3} \int \frac{d t}{t^{2}} = \frac{4}{3 t} + C = \frac{4}{3} (\frac{x - 1}{x + 2})^{\frac{1}{4}} + C,$ kur $((x - 1)^{3} (x + 2)^{5})^{\frac{1}{4}} = (x - 1) (x + 2) (\frac{x + 2}{x - 1})^{\frac{1}{4}};$ $\frac{x + 2}{x - 1} = t^{4};$ $x = \frac{t^{4} + 2}{t^{4} - 1};$ $x - 1 = \frac{3}{t^{4} - 1};$ $x + 2 = \frac{3 t^{4}}{t^{4} - 1};$ $d x = \frac{- 12 t^{3}}{(t^{4} - 1)^{2}} d t .$

$\int \frac{1}{(1 - x)^{2}} (\frac{1 + x}{1 - x})^{\frac{1}{3}} d x = \int \frac{(1 + u^{3})^{2} u \cdot 6 u^{2}}{4 (1 + u^{3})^{2}} d u = \frac{3}{2} \cdot \frac{u^{4}}{4} + C = \frac{3}{8} \cdot \frac{1 + x}{1 - x} (\frac{1 + x}{1 - x})^{\frac{1}{3}} + C,$ kur $\frac{1 + x}{1 - x} = u^{3};$ $x = \frac{u^{3} - 1}{u^{3} + 1};$ $d x = \frac{6 u^{2}}{(1 + u^{3})^{2}} d u;$ $1 - x = \frac{2}{1 + u^{3}} .$

Papildomai

Iracionaliųjų funkcijų integravimas (papildomai)

Oilerio keitiniai

Funkcijos

R (x, \sqrt{a x^{2} + b x + c}) (7.66)

(a, b ir c - realieji skačiai) integralas yra elementarioji funkcija.

I. $\sqrt{a x^{2} + b x + c} = u \pm x \sqrt{a}$ ; a>0;

u = \sqrt{a x^{2} + b x + c} + x \sqrt{a}

. Pakėlus abi dalis lygybės

\sqrt{a x^{2} + b x + c} = u - x \sqrt{a}

kvadartu, gauname

b x + c = u^{2} - 2 \sqrt{a} u x,

taip kad

x = \frac{u^{2} - c}{2 \sqrt{a} u + b},

\sqrt{a x^{2} + b x + c} = u - \frac{u^{2} - c}{2 \sqrt{a} u + b} \cdot \sqrt{a} = \frac{u (2 \sqrt{a} u + b) - (\sqrt{a} u^{2} - \sqrt{a} c)}{2 \sqrt{a} u + b} = \frac{\sqrt{a} u^{2} + b u + c \sqrt{a}}{2 \sqrt{a} u + b},

d x = (\frac{u^{2} - c}{2 \sqrt{a} u + b})^{'} = \frac{2 u \cdot (2 \sqrt{a} u + b) - (u^{2} - c) \cdot 2 \sqrt{a}}{(2 \sqrt{a} u + b)^{2}} d u = 2 \frac{\sqrt{a} u^{2} + b u + c \sqrt{a}}{(2 \sqrt{a} u + b)^{2}} d u .

Tinka, kai kvadrainis trinaris turi menamas šaknis.

II. $\sqrt{a x^{2} + b x + c} = x t \pm \sqrt{c}$ ; c>0. Tada

a x^{2} + b x + c = x^{2} t^{2} + 2 x t \sqrt{c} + c .

(Mes apsisprendėme prieš šaknį pasirinkti pliuso ženklą.) Iš čia x yra kaip racionali funkcija nuo t:

a x^{2} - x^{2} t^{2} + b x - 2 x t \sqrt{c} = 0,

(a - t^{2}) x^{2} + (b - 2 t \sqrt{c}) x = 0,

x ((a - t^{2}) x + b - 2 t \sqrt{c}) = 0;

iš čia arba x=0 arba

(a - t^{2}) x + b - 2 t \sqrt{c} = 0;

Mums reikia išreikšti x per t, todėl x=0 netinka; sprendžiame toliau:

(a - t^{2}) x + b - 2 t \sqrt{c} = 0;

(a - t^{2}) x = 2 \sqrt{c} t - b;

x = \frac{2 \sqrt{c} t - b}{a - t^{2}} .

Turime

\sqrt{a x^{2} + b x + c} = x t + \sqrt{c} = \frac{2 \sqrt{c} t - b}{a - t^{2}} \cdot t + \sqrt{c} = \frac{2 \sqrt{c} t^{2} - b t}{a - t^{2}} + \sqrt{c};

(\frac{2 \sqrt{c} t - b}{a - t^{2}}) = \frac{(2 \sqrt{c} t - b)^{'} (a - t^{2}) - (2 \sqrt{c} t - b) (a - t^{2})^{'}}{(a - t^{2})^{2}} = \frac{2 \sqrt{c} (a - t^{2}) - (2 \sqrt{c} t - b) (- 2 t)}{(a - t^{2})^{2}} = \frac{2 a \sqrt{c} - 2 \sqrt{c} t^{2} - (- 4 \sqrt{c} t^{2} + 2 b t)}{(a - t^{2})^{2}} =

= \frac{2 a \sqrt{c} - 2 \sqrt{c} t^{2} + 4 \sqrt{c} t^{2} - 2 b t}{(a - t^{2})^{2}} = \frac{(4 \sqrt{c} - 2 \sqrt{c}) t^{2} - 2 b t + 2 a \sqrt{c}}{(a - t^{2})^{2}} .

d x = \frac{2 \sqrt{c} t^{2} - 2 b t + 2 a \sqrt{c}}{(a - t^{2})^{2}} d t .

III. $\sqrt{a (x - x_{1}) (x - x_{2})} = u (x - x_{1}),$ kur $x_{1}$ yra bet kuri realioji trinario $a x^{2} + b x + c$ šaknis. Taikoma tik kai yra du lygties $\sqrt{a x^{2} + b x + c} = \sqrt{a (x - x_{1}) (x - x_{2})}$ sprendiniai.

Tegu

α

ir

β

- realiosios šaknys trinario

a x^{2} + b x + c .

Tada

\sqrt{a x^{2} + b x + c} = (x - α) t .

Kadangi

a x^{2} + b x + c = a (x - α) (x - β),

tai

\sqrt{a (x - α) (x - β)} = (x - α) t,

a (x - α) (x - β) = (x - α)^{2} t^{2},

a (x - β) = (x - α) t^{2} .

Iš čia randame x kaip racionalią funkciją nuo t:

a x - a β = x t^{2} - α t^{2},

a x - x t^{2} = a β - α t^{2},

(a - t^{2}) x = a β - α t^{2},

x = \frac{a β - α t^{2}}{a - t^{2}} .

Trečias Oilerio keitinys tinka kai a>0 ir kai a<0. Butina tik, kad turėtų daugianaris

a x^{2} + b x + c

dvi realiasias šaknis.

(\frac{a β - α t^{2}}{a - t^{2}})^{'} = \frac{- 2 α t (a - t^{2}) - (a β - α t^{2}) (- 2 t)}{(a - t^{2})^{2}} = \frac{- 2 α a t + 2 α t^{3} - (- 2 a β t + 2 α t^{3})}{(a - t^{2})^{2}} =

= \frac{- 2 α a t + 2 α t^{3} + 2 a β t - 2 α t^{3}}{(a - t^{2})^{2}} = \frac{- 2 α a t + 2 a β t}{(a - t^{2})^{2}} = \frac{(2 a β - 2 a α) t}{(a - t^{2})^{2}} .

d x = \frac{2 a (β - α) t}{(a - t^{2})^{2}} d t .

Pavyzdžiai

Pirmojo Oilerio keitinio pavyzdžiai

$\int \frac{d x}{\sqrt{5 x^{2} + x + 1}};$

$\sqrt{5 x^{2} + x + 1} = u + x \sqrt{5} .$ Pakėlę šios lygybės abi puses kvadratu, gauname: $5 x^{2} + x + 1 = u^{2} + 2 u x \sqrt{5} + 5 x^{2}; x + 1 = u^{2} + 2 \sqrt{5} u x; x (1 - 2 u \sqrt{5}) = u^{2} - 1; x = \frac{u^{2} - 1}{1 - 2 \sqrt{5} u};$

d x = (\frac{u^{2} - 1}{1 - 2 \sqrt{5} u})^{'} = \frac{(u^{2} - 1)^{'} (1 - 2 \sqrt{5} u) - (u^{2} - 1) (1 - 2 \sqrt{5} u)^{'}}{(1 - 2 \sqrt{5} u)^{2}} =

= \frac{2 u (1 - 2 \sqrt{5} u) - (u^{2} - 1) (- 2 \sqrt{5})}{1 - 4 \sqrt{5} u + 4 \cdot 5 \cdot u^{2}} d u = \frac{2 u - 4 \sqrt{5} u^{2} + 2 \sqrt{5} (u^{2} - 1)}{1 - 4 \sqrt{5} u + 20 u^{2}} d u =

= \frac{2 u - 4 \sqrt{5} u^{2} + 2 \sqrt{5} \cdot u^{2} - 2 \sqrt{5}}{1 - 4 \sqrt{5} u + 20 u^{2}} d u = \frac{2 u - 2 \sqrt{5} u^{2} - 2 \sqrt{5}}{1 - 4 \sqrt{5} u + 20 u^{2}} d u = \frac{2 (u - \sqrt{5} u^{2} - \sqrt{5})}{(1 - 2 \sqrt{5} \cdot u)^{2}} d u .

\sqrt{5 x^{2} + x + 1} = u + x \sqrt{5} = u + \frac{u^{2} - 1}{1 - 2 \sqrt{5} u} \cdot \sqrt{5} = u + \frac{\sqrt{5} u^{2} - \sqrt{5}}{1 - 2 \sqrt{5} u} = \frac{u (1 - 2 \sqrt{5} u) + \sqrt{5} u^{2} - \sqrt{5}}{1 - 2 \sqrt{5} u} =

= \frac{u - 2 \sqrt{5} u^{2} + \sqrt{5} u^{2} - \sqrt{5}}{1 - 2 \sqrt{5} u} = \frac{u - \sqrt{5} u^{2} - \sqrt{5}}{1 - 2 \sqrt{5} u} .

\int \frac{d x}{\sqrt{5 x^{2} + x + 1}} = \int \frac{\frac{2 (u - \sqrt{5} u^{2} - \sqrt{5})}{(1 - 2 \sqrt{5} \cdot u)^{2}} d t}{\frac{u - \sqrt{5} u^{2} - \sqrt{5}}{1 - 2 \sqrt{5} u}} = 2 \int \frac{d u}{1 - 2 \sqrt{5} u} = - \frac{2}{2 \sqrt{5}} \int \frac{d (1 - u 2 \sqrt{5})}{1 - 2 \sqrt{5} u} =

$= - \frac{1}{\sqrt{5}} \ln | 1 - 2 \sqrt{5} u | + C = - \frac{1}{\sqrt{5}} \ln | 1 - 2 \sqrt{5} (\sqrt{5 x^{2} + x + 1} - \sqrt{5} x) | + C =$ $= - \frac{1}{\sqrt{5}} \ln | 1 - \sqrt{100 x^{2} + 20 x + 20} + 10 x | + C .$

$\int \frac{d x}{\sqrt{x^{2} + 3}} .$ Taikome I Oilerio keitinį $\sqrt{x^{2} + 3} = - x + t = t - x .$

$x^{2} + 3 = (- x + t)^{2} = x^{2} - 2 t x + t^{2}$ ; $3 = t^{2} - 2 t x$ ; $2 t x = t^{2} - 3$ ; $x = \frac{t^{2} - 3}{2 t} .$ $d x = (\frac{t^{2} - 3}{2 t})^{'} = \frac{(t^{2} - 3)^{'} \cdot 2 t - (t^{2} - 3) \cdot (2 t)^{'}}{(2 t)^{2}} = \frac{2 t \cdot 2 t - (t^{2} - 3) \cdot 2}{4 t^{2}} d t = \frac{2 t^{2} - (t^{2} - 3)}{2 t^{2}} d t = \frac{t^{2} + 3}{2 t^{2}} d t .$

\sqrt{x^{2} + 3} = - x + t = - \frac{t^{2} - 3}{2 t} + t = \frac{- (t^{2} - 3) + 2 t^{2}}{2 t} = \frac{t^{2} + 3}{2 t} .

$\int \frac{d x}{\sqrt{x^{2} + 3}} = \int \frac{\frac{t^{2} + 3}{2 t^{2}} d t}{\frac{t^{2} + 3}{2 t}} = \int \frac{d t}{t} = \ln | t | + C = \ln | x + \sqrt{x^{2} + 3} | + C .$

Apskaičiuoti $\int \frac{d x}{x + \sqrt{x^{2} + x + 1}} .$

Sprendimas. Kadangi trinaris

x^{2} + x + 1

turi kompleksines šaknis, padarysime keitinį

\sqrt{x^{2} + x + 1} = t - x .

Pakėlę abi lygybės puses kvadratu, gauname

$x^{2} + x + 1 = t^{2} - 2 t x + x^{2}$ arba $x + 1 = t^{2} - 2 t x$ ; iš čia $x = \frac{t^{2} - 1}{1 + 2 t}$ , $d x = (\frac{t^{2} - 1}{1 + 2 t})^{'} = \frac{2 t (1 + 2 t) - (t^{2} - 1) \cdot 2}{(1 + 2 t)^{2}} = \frac{2 t + 4 t^{2} - 2 t^{2} + 2}{(1 + 2 t)^{2}} = \frac{2 t + 2 t^{2} + 2}{(1 + 2 t)^{2}} .$

Tada

$\int \frac{d x}{x + \sqrt{x^{2} + x + 1}} = \int \frac{2 t + 2 t^{2} + 2}{t (1 + 2 t)^{2}} d t .$

Toliau, turime

$\frac{2 t^{2} + 2 t + 2}{t (1 + 2 t)^{2}} = \frac{A}{t} + \frac{B}{1 + 2 t} + \frac{D}{(1 + 2 t)^{2}} .$

Padauginę abi dalis lygybės su

t (1 + 2 t)^{2},

gauname

2 t^{2} + 2 t + 2 = A (1 + 2 t)^{2} + B t (1 + 2 t) + D t = A (1 + 4 t + 4 t^{2}) + B t + 2 B t^{2} + D t = (4 A + 2 B) t^{2} + (4 A + B + D) t + A .

Prilyginę koeficientus prie vienodų laipsmių t, gauname sistemą lygčių pirmojo laipsnio atžvilgiu A, B, D:

{4A+2B=2,

{4A+B+D=2,

{A=2,

Iš čia A=2, B=-3, D=-3. Todėl,

\frac{2 t^{2} + 2 t + 2}{t (1 + 2 t)^{2}} = \frac{2}{t} - \frac{3}{1 + 2 t} - \frac{3}{(1 + 2 t)^{2}},

ir galutinai

\int \frac{d x}{x + \sqrt{x^{2} + x + 1}} = \int [\frac{2}{t} - \frac{3}{1 + 2 t} - \frac{3}{(1 + 2 t)^{2}}] d t = 2 \int \frac{d t}{t} - \frac{3}{2} \int \frac{d (1 + 2 t)}{1 + 2 t} - \frac{3}{2} \int \frac{d (1 + 2 t)}{(1 + 2 t)^{2}} =

= 2 \ln | t | - \frac{3}{2} \ln | 1 + 2 t | + \frac{3}{2 (1 + 2 t)} + C =

$= 2 \ln | x + \sqrt{x^{2} + x + 1} | - \frac{3}{2} \ln | 1 + 2 x + 2 \sqrt{x^{2} + x + 1} | + \frac{3}{2 + 4 x + 4 \sqrt{x^{2} + x + 1}} + C .$

Antrojo Oilerio keitinio pavyzdžiai

$\int \frac{d x}{\sqrt{1 - 6 x - 9 x^{2}}};$

$\sqrt{1 - 6 x - 9 x^{2}} = u x + 1 .$ Pakelę abi puses kvadratu, gauname $1 - 6 x - 9 x^{2} = u^{2} x^{2} + 2 u x + 1;$ $- 6 x - 9 x^{2} = u^{2} x^{2} + 2 u x;$ $- 6 - 9 x = u^{2} x + 2 u .$ Imdami apiejų lygybės pusių diferencialus, randame: $- 9 d x = 2 u x d u + 2 d u;$ $d x = \frac{- 2 (u x + 1)}{9 + u^{2}} d u .$ $\int \frac{d x}{\sqrt{1 - 6 x - 9 x^{2}}} = - 2 \int \frac{(u x + 1) d u}{(9 + u^{2}) (u x + 1)} = - 2 \int \frac{d u}{9 + u^{2}} = - \frac{2}{3} \arctan \frac{u}{3} + C =$ $= - \frac{2}{3} \arctan \frac{\sqrt{1 - 6 x - 9 x^{2}} - 1}{3 x} + C .$

Sprendimas normaliai. Čia a=-9, b=-6, c=1. Tada

x = \frac{2 \sqrt{c} t - b}{a - t^{2}} = \frac{2 \sqrt{1} t - (- 6)}{- 9 - t^{2}} = \frac{2 t + 6}{- 9 - t^{2}} .

\sqrt{1 - 6 x - 9 x^{2}} = x t + 1 = \frac{2 t + 6}{- 9 - t^{2}} t + 1 = \frac{2 t^{2} + 6 t}{- 9 - t^{2}} + 1 = \frac{2 t^{2} + 6 t - 9 - t^{2}}{- 9 - t^{2}} = \frac{t^{2} + 6 t - 9}{- 9 - t^{2}} .

d x = \frac{2 \sqrt{c} t^{2} - 2 b t + 2 a \sqrt{c}}{(a - t^{2})^{2}} d t = \frac{2 \sqrt{1} t^{2} - 2 (- 6) t + 2 (- 9) \sqrt{1}}{(- 9 - t^{2})^{2}} d t = \frac{2 t^{2} + 12 t - 18}{(- 9 - t^{2})^{2}} d t = \frac{2 (t^{2} + 6 t - 9)}{(- 9 - t^{2})^{2}} d t .

\int \frac{d x}{\sqrt{1 - 6 x - 9 x^{2}}} = \int \frac{\frac{2 (t^{2} + 6 t - 9)}{(- 9 - t^{2})^{2}} d t}{\frac{t^{2} + 6 t - 9}{- 9 - t^{2}}} = \int \frac{\frac{2}{- 9 - t^{2}} d t}{1} = \int \frac{2 d t}{- 9 - t^{2}} = - 2 \int \frac{d t}{9 + t^{2}} = - \frac{2}{3} \arctan \frac{t}{3} + C =

= - \frac{2}{3} \arctan \frac{\sqrt{1 - 6 x - 9 x^{2}} - 1}{3 x} + C .

Apskaičiuoti $\int \frac{d x}{(1 + x) \sqrt{1 + x - x^{2}}} .$

Sprendimas. Čia trinaris

1 + x - x^{2}

turi kompleksines šaknis ir a<0, c>0, pasinaudojame keitiniu

\sqrt{1 + x - x^{2}} = t x - 1 .

Pakėlę abi lygties puses kvadratu, gauname

$1 + x - x^{2} = t^{2} x^{2} - 2 t x + 1$ , $x - x^{2} = t^{2} x^{2} - 2 t x$ , $1 - x = t^{2} x - 2 t$ , $1 + 2 t = t^{2} x + x$ , $x (t^{2} + 1) = 1 + 2 t$ , $x = \frac{1 + 2 t}{t^{2} + 1}$ . $d x = (\frac{1 + 2 t}{t^{2} + 1})^{'} = \frac{2 (t^{2} + 1) - (1 + 2 t) \cdot 2 t}{(t^{2} + 1)^{2}} d t = \frac{2 t^{2} + 2 - 2 t - 4 t^{2}}{(t^{2} + 1)^{2}} d t = \frac{2 - 2 t - 2 t^{2}}{(t^{2} + 1)^{2}} d t = \frac{2 (1 - t - t^{2})}{(t^{2} + 1)^{2}} d t .$ $\sqrt{1 + x - x^{2}} = t \cdot \frac{1 + 2 t}{t^{2} + 1} - 1 = \frac{t + 2 t^{2} - t^{2} - 1}{t^{2} + 1} = \frac{t + t^{2} - 1}{t^{2} + 1} .$ $\sqrt{1 + x - x^{2}} = t x - 1; \sqrt{1 + x - x^{2}} + 1 = t x; t = \frac{\sqrt{1 + x - x^{2}} + 1}{x} .$

Tokiu budu,

$\int \frac{d x}{(1 + x) \sqrt{1 + x - x^{2}}} = \int \frac{\frac{2 (1 - t - t^{2})}{(t^{2} + 1)^{2}} d t}{(1 + \frac{1 + 2 t}{t^{2} + 1}) \frac{t + t^{2} - 1}{t^{2} + 1}} = \int \frac{\frac{- 2 (t^{2} + t - 1)}{(t^{2} + 1)^{2}} d t}{\frac{t^{2} + 1 + 1 + 2 t}{t^{2} + 1} \cdot \frac{t^{2} + t - 1}{t^{2} + 1}} = \int \frac{- 2 d t}{1 + t^{2} + 2 t + 1} =$ $= \int \frac{- 2 d (t + 1)}{1 + (t + 1)^{2}} = - 2 \arctan (t + 1) + C = - 2 \arctan (\frac{\sqrt{1 + x - x^{2}} + 1}{x} + 1) + C =$ $= - 2 \arctan \frac{\sqrt{1 + x - x^{2}} + 1 + x}{x} + C .$

Reikia apskaičiuoti integralą

$\int \frac{(1 - \sqrt{1 + x + x^{2}})^{2}}{x^{2} \sqrt{1 + x + x^{2}}} d x .$

Sprendimas. Taikome II Oilerio keitinį

\sqrt{1 + x + x^{2}} = x t + 1;

tada

1 + x + x^{2} = x^{2} t^{2} + 2 x t + 1;

x + x^{2} = x^{2} t^{2} + 2 x t;

1 + x = x t^{2} + 2 t;

x - x t^{2} = 2 t - 1;

x (1 - t^{2}) = 2 t - 1;

x = \frac{2 t - 1}{1 - t^{2}} .

$d x = \frac{2 (1 - t^{2}) - (2 t - 1) \cdot (- 2 t)}{(1 - t^{2})^{2}} d t = \frac{2 - 2 t^{2} + 2 t (2 t - 1)}{(1 - t^{2})^{2}} d t = \frac{2 - 2 t^{2} + 4 t^{2} - 2 t}{(1 - t^{2})^{2}} d t = \frac{2 t^{2} - 2 t + 2}{(1 - t^{2})^{2}} d t .$

\sqrt{1 + x + x^{2}} = \frac{2 t - 1}{1 - t^{2}} \cdot t + 1 = \frac{2 t^{2} - t + (1 - t^{2})}{1 - t^{2}} = \frac{t^{2} - t + 1}{1 - t^{2}} .

1 - \sqrt{1 + x + x^{2}} = 1 - (x t + 1) = - x t = - \frac{2 t - 1}{1 - t^{2}} \cdot t = \frac{- 2 t^{2} + t}{1 - t^{2}} .

t = \frac{\sqrt{1 + x + x^{2}} - 1}{x} .

Įstate gautas išraiškas į pradinį integralą, randame:

$\int \frac{(1 - \sqrt{1 + x + x^{2}})^{2}}{x^{2} \sqrt{1 + x + x^{2}}} d x = \int \frac{(- 2 t^{2} + t)^{2} (1 - t^{2})^{2} (1 - t^{2}) (2 t^{2} - 2 t + 2)}{(1 - t^{2})^{2} (2 t - 1)^{2} (t^{2} - t + 1) (1 - t^{2})^{2}} d t = 2 \int \frac{(- 2 t^{2} + t)^{2}}{(2 t - 1)^{2} (1 - t^{2})} d t = 2 \int \frac{t^{2}}{1 - t^{2}} d t =$ $= 2 \int (\frac{1}{1 - t^{2}} - 1) d t = 2 (\frac{1}{2} \ln | \frac{1 + t}{1 - t} | - t) + C =$ $= \ln | \frac{1 + \frac{\sqrt{1 + x + x^{2}} - 1}{x}}{1 - \frac{\sqrt{1 + x + x^{2}} - 1}{x}} | - \frac{2 (\sqrt{1 + x + x^{2}} - 1)}{x} + C = \ln | \frac{\frac{x + \sqrt{1 + x + x^{2}} - 1}{x}}{\frac{x - \sqrt{1 + x + x^{2}} + 1}{x}} | - \frac{2 (\sqrt{1 + x + x^{2}} - 1)}{x} + C =$ $= \ln | \frac{x + \sqrt{1 + x + x^{2}} - 1}{x - \sqrt{1 + x + x^{2}} + 1} | - \frac{2 (\sqrt{1 + x + x^{2}} - 1)}{x} + C =$ $= \ln | 2 x + 2 \sqrt{1 + x + x^{2}} + 1 | - \frac{2 (\sqrt{1 + x + x^{2}} - 1)}{x} + C .$

Trečiojo Oilerio keitinio pavyzdžiai

$\int \frac{d x}{x \sqrt{- x^{2} + 5 x - 6}} .$ Pastebėję, kad pošaknio trinario šaknys yra 2 ir 3, taikome keitinį $\sqrt{(x - 2) (3 - x)} = u (x - 2) .$ Pakėlę šios lygybės abi puses kvadratu ir suprastinę iš $(x - 2),$ gauname:

(x - 2) (3 - x) = u^{2} (x - 2)^{2};

3 - x = u^{2} (x - 2);

- x u^{2} - x = - 2 u^{2} - 3;

x (- u^{2} - 1) = - 2 u^{2} - 3;

x = \frac{2 u^{2} + 3}{u^{2} + 1}; u = \sqrt{\frac{3 - x}{x - 2}}

;

d x = \frac{4 u (u^{2} + 1) - 2 u (2 u^{2} + 3)}{(u^{2} + 1)^{2}} d u = \frac{4 u^{3} + 4 u - 4 u^{3} - 6 u}{(u^{2} + 1)^{2}} d u = \frac{- 2 u d u}{(u^{2} + 1)^{2}} .

\sqrt{(x - 2) (3 - x)} = u (x - 2) = u (\frac{2 u^{2} + 3}{u^{2} + 1} - 2) = \frac{2 u^{3} + 3 u - 2 u (u^{2} + 1)}{u^{2} + 1} = \frac{2 u^{3} + 3 u - 2 u^{3} - 2 u}{u^{2} + 1} = \frac{u}{u^{2} + 1} .

\int \frac{d x}{x \sqrt{- x^{2} + 5 x - 6}} = \int \frac{\frac{- 2 u d u}{(u^{2} + 1)^{2}}}{\frac{2 u^{2} + 3}{u^{2} + 1} \cdot \frac{u}{u^{2} + 1}} = - 2 \int \frac{d u}{3 + 2 u^{2}} = - \int \frac{d u}{\frac{3}{2} + u^{2}} =

= - \frac{1}{\sqrt{\frac{3}{2}}} \arctan (\frac{u}{\sqrt{\frac{3}{2}}}) + C = - \sqrt{\frac{2}{3}} \arctan (\sqrt{\frac{2}{3}} u) + C = - \sqrt{\frac{2}{3}} \arctan (\sqrt{\frac{2}{3}} \cdot \sqrt{\frac{3 - x}{x - 2}}) + C .

$\int \frac{x d x}{\sqrt{(7 x - 10 - x^{2})^{3}}},$ kur $a < 0,$ $c < 0,$ todėl taikome III Oilerio keitinį. Lygties $7 x - 10 - x^{2}$ sprendiniai yra $x_{1} = 2$ , $x_{2} = 5$ ; $a = - 1 .$

$\sqrt{7 x - 10 - x^{2}} = \sqrt{- 1 (x - 2) (x - 5)} = \sqrt{(x - 2) (5 - x)} = (x - 2) t .$ $5 - x = (x - 2) t^{2};$ $- x t^{2} - x = - 2 t^{2} - 5;$ $x = \frac{5 + 2 t^{2}}{1 + t^{2}};$ $d x = (\frac{5 + 2 t^{2}}{1 + t^{2}})^{'} = \frac{4 t (1 + t^{2}) - (5 + 2 t^{2}) 2 t}{(1 + t^{2})^{2}} d t = \frac{4 t + 4 t^{3} - 10 t - 4 t^{3}}{(1 + t^{2})^{2}} d t = \frac{- 6 t}{(1 + t^{2})^{2}} d t .$ $\sqrt{7 x - 10 - x^{2}} = (x - 2) t = (\frac{5 + 2 t^{2}}{1 + t^{2}} - 2) t = \frac{5 + 2 t^{2} - 2 - 2 t^{2}}{1 + t^{2}} t = \frac{3 t}{1 + t^{2}} .$ $t = \frac{\sqrt{(x - 2) (5 - x)}}{(x - 2)} = \sqrt{\frac{(5 - x)}{(x - 2)}} .$ $\int \frac{x d x}{\sqrt{(7 x - 10 - x^{2})^{3}}} = \int \frac{\frac{5 + 2 t^{2}}{1 + t^{2}} \frac{- 6 t}{(1 + t^{2})^{2}} d t}{(\frac{3 t}{1 + t^{2}})^{3}} = \int \frac{\frac{- 6 t (5 + 2 t^{2})}{(1 + t^{2})^{3}}}{(\frac{3 t}{1 + t^{2}})^{3}} d t = \int \frac{- 6 t (5 + 2 t^{2})}{27 t^{3}} d t = \int \frac{- 6 (5 + 2 t^{2})}{27 t^{2}} d t =$ $= \int (- \frac{30}{27 t^{2}} - \frac{12 t^{2}}{27 t^{2}}) d t = \frac{1}{27} \int (- \frac{30}{t^{2}} - 12) d t = \frac{1}{27} (\frac{30}{t} - 12 t) + C = \frac{1}{27} (\frac{30}{\sqrt{\frac{(5 - x)}{(x - 2)}}} - 12 \sqrt{\frac{(5 - x)}{(x - 2)}}) + C =$ $= \frac{30 \sqrt{x - 2}}{27 \sqrt{5 - x}} - \frac{12}{27} \sqrt{\frac{(5 - x)}{(x - 2)}} + C .$

Reikia apskaičiuoti integralą

$\int \frac{d x}{\sqrt{x^{2} + 3 x - 4}} .$

Sprendimas. Kadangi

x^{2} + 3 x - 4 = (x + 4) (x - 1),

tai:

$\sqrt{(x + 4) (x - 1)} = (x + 4) t;$ tada

(x + 4) (x - 1) = (x + 4)^{2} t^{2}

,

x - 1 = (x + 4) t^{2},

x - 1 = x t^{2} + 4 t^{2},

x - x t^{2} = 4 t^{2} + 1,

x = \frac{1 + 4 t^{2}}{1 - t^{2}} .

d x = \frac{(1 + 4 t^{2})^{'} \cdot (1 - t^{2}) - (1 + 4 t^{2}) \cdot (1 - t^{2})^{'}}{(1 - t^{2})^{2}} = \frac{8 t \cdot (1 - t^{2}) - (1 + 4 t^{2}) \cdot (- 2 t)}{(1 - t^{2})^{2}} d t =

= \frac{8 t - 8 t^{3} + 2 t + 8 t^{3}}{(1 - t^{2})^{2}} d t = \frac{10 t}{(1 - t^{2})^{2}} d t .

\sqrt{(x + 4) (x - 1)} = (x + 4) t = (\frac{1 + 4 t^{2}}{1 - t^{2}} + 4) t = \frac{1 + 4 t^{2} + 4 - 4 t^{2}}{1 - t^{2}} \cdot t = \frac{5 t}{1 - t^{2}} .

t = \frac{\sqrt{(x + 4) (x - 1)}}{(x + 4)} = \sqrt{\frac{(x + 4) (x - 1)}{(x + 4)^{2}}} = \sqrt{\frac{x - 1}{x + 4}} .

Grįžtant prie pradinio integralo, gauname:

$\int \frac{d x}{\sqrt{x^{2} + 3 x - 4}} = \int \frac{10 t (1 - t^{2})}{(1 - t^{2})^{2} 5 t} d t = \int \frac{2}{1 - t^{2}} d t = \ln | \frac{1 + t}{1 - t} | + C = \ln | \frac{1 + \sqrt{\frac{x - 1}{x + 4}}}{1 - \sqrt{\frac{x - 1}{x + 4}}} | + C =$ $= \ln | \frac{\sqrt{x + 4} + \sqrt{x - 1}}{\sqrt{x + 4} - \sqrt{x - 1}} | + C .$

Diferencialinių binomų integravimas

Integralas $\int x^{m} (a + b x^{n})^{p} d x,$ kur m, n, p - racionalieji skaičiai, vadinamas integralu su binominiu diferencialu. Šį integralą elementariosiomis funkcijomis įmanoma išreikšti tik trimis atvejais:

I. p - sveikasis skaičius. Jei

p > 0,

tai pointegralinis binomas skleidžiamas pagal Niutono binomo formulę. Jei

p < 0,

tai keičiame

x = t^{k},

kur k - bendras trupmenų m ir n vardiklis. Pavyzdžiui, trupmenų

\frac{1}{4}

ir

\frac{2}{3}

bendras vardiklis yra 3

\cdot

4 = 12.

II.

\frac{m + 1}{n}

- sveikasis skaičius. Keičiame

a + b x^{n} = t^{α},

kur

α

- trupmenos p vardiklis.

III.

\frac{m + 1}{n} + p

- sveikasis skaičius. Keičiame

a + b x^{n} = t^{α} x^{n},

kur

α

- trupmenos p vardiklis.

Pavyzdžiai

$\int x^{\frac{1}{3}} (2 + \sqrt{x})^{2} d x = \int x^{\frac{1}{3}} (4 + 4 \sqrt{x} + x) d x = \int (4 x^{\frac{1}{3}} + 4 x^{\frac{5}{6}} + x^{\frac{4}{3}}) d x = 3 x^{\frac{4}{3}} + \frac{24}{11} x^{\frac{11}{6}} + \frac{3}{7} x^{\frac{7}{3}} + C,$

kur $p = 2$ - sveikasis skaičius. Turime I atvejį.

$\int \frac{\sqrt{1 + x^{\frac{1}{3}}}}{x^{\frac{2}{3}}} d x = \int x^{- \frac{2}{3}} (1 + x^{\frac{1}{3}})^{\frac{1}{2}} d x = \int \sqrt{t^{2}} 6 t d t = 6 \int t^{2} d t = 2 t^{3} + C = 2 (1 + x^{\frac{1}{3}})^{\frac{3}{2}} + C,$

kur $m = - \frac{2}{3}; n = \frac{1}{3}; p = \frac{1}{2}; \frac{m + 1}{n} = 1 .$ Turime II atvejį. $1 + x^{\frac{1}{3}} = t^{2}; \frac{1}{3} x^{- \frac{2}{3}} d x = 2 t d t .$

$\int x^{- 11} (1 + x^{4})^{- \frac{1}{2}} d x = - \frac{1}{2} \int (t^{2} - 1)^{\frac{11}{4}} (\frac{t^{2}}{t^{2} - 1})^{- \frac{1}{2}} \frac{t d t}{(t^{2} - 1)^{\frac{5}{4}}} = - \frac{1}{2} \int (t^{2} - 1)^{\frac{11}{4} + \frac{1}{2} - \frac{5}{4}} \frac{t}{t^{\frac{2}{2}}} d t =$

$= - \frac{1}{2} \int (t^{2} - 1)^{2} d t = - \frac{t^{5}}{10} + \frac{t^{3}}{3} - \frac{t}{2} + C = - \frac{(1 + x^{4})^{\frac{5}{2}}}{10 x^{2}} + \frac{(1 + x^{4})^{\frac{3}{2}}}{3 x^{2}} - \frac{(1 + x^{4})^{\frac{1}{2}}}{2 x^{2}} + C,$ kur $m = - 11;$ $n = 4;$ $p = - \frac{1}{2}; \frac{m + 1}{n} + p = - 3 .$ Turime III atvejį. $1 + x^{4} = x^{4} t^{2};$ $x = \frac{1}{(t^{2} - 1)^{\frac{1}{4}}}; d x = - \frac{t d t}{2 (t^{2} - 1)^{\frac{5}{4}}}; t = \frac{\sqrt{1 + x^{4}}}{x^{2}} .$

$\int \frac{x^{3} d x}{\sqrt{1 + 2 x^{2}}} = \int x^{3} (1 + 2 x^{2})^{- \frac{1}{2}} d x = \frac{1}{2} \int (\frac{u^{2} - 1}{2})^{\frac{3}{2}} u^{- 1} (\frac{u^{2} - 1}{2})^{- \frac{1}{2}} u d u = \frac{1}{2} \int \frac{u^{2} - 1}{2} d u =$

$= \frac{1}{4} (\frac{u^{3}}{3} - u) + C = \frac{(1 + 2 x^{2})^{\frac{3}{2}}}{12} - \frac{\sqrt{1 + 2 x^{2}}}{4} + C,$ kur $\frac{m + 1}{n} = \frac{3 + 1}{2} = 2;$ $1 + 2 x^{2} = u^{2};$ $x = (\frac{u^{2} - 1}{2})^{\frac{1}{2}}; d x = \frac{1}{2} (\frac{u^{2} - 1}{2})^{- \frac{1}{2}} u d u .$

$\int x^{- \frac{1}{3}} (1 - x^{\frac{2}{3}})^{- \frac{1}{2}} d x = - 3 \int t^{- 1} \cdot t d t = - 3 \int d t = - 3 t + C = - 3 \sqrt{1 - x^{\frac{2}{3}}} + C,$

kur $m = - \frac{1}{3};$ $n = \frac{2}{3}; p = - \frac{1}{2}; \frac{m + 1}{n} = 1; 1 - x^{\frac{2}{3}} = t^{2}; - \frac{2}{3} x^{- \frac{1}{3}} d x = 2 t d t; d x = - 3 x^{\frac{1}{3}} t d t; x^{\frac{2}{3}} = 1 - t^{2} .$

$\int x^{\frac{1}{3}} (1 - x^{\frac{2}{3}})^{- \frac{1}{2}} d x = - 3 \int x^{\frac{1}{3}} \cdot (t^{2})^{- \frac{1}{2}} \cdot x^{\frac{1}{3}} \cdot t d t = - 3 \int x^{\frac{2}{3}} \cdot t^{- 1} \cdot t d t =$

$= - 3 \int (1 - t^{2}) \frac{t}{t} d t = - 3 \int (1 - t^{2}) d t = - 3 (t - \frac{t^{3}}{3}) + C = - 3 (\sqrt{1 - x^{\frac{2}{3}}} - \frac{\sqrt{(1 - x^{\frac{2}{3}})^{3}}}{3}) + C =$ $= - 3 \frac{3 (1 - x^{\frac{2}{3}})^{\frac{1}{2}} - (1 - x^{\frac{2}{3}})^{\frac{3}{2}}}{3} + C = - 3 (1 - x^{\frac{2}{3}})^{\frac{1}{2}} + (1 - x^{\frac{2}{3}})^{\frac{3}{2}} + C =$

= - 3 \sqrt{1 - x^{\frac{2}{3}}} + \sqrt{1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}} + C,

kur $m = \frac{1}{3};$ $n = \frac{2}{3}; p = - \frac{1}{2}; \frac{m + 1}{n} = 2; 1 - x^{\frac{2}{3}} = t^{2}; - \frac{2}{3} x^{- \frac{1}{3}} d x = 2 t d t; d x = - 3 x^{\frac{1}{3}} t d t; x^{\frac{2}{3}} = 1 - t^{2} .$

$\int \frac{d x}{\sqrt{x^{2} + 3}} .$ Matome, kad tinka trečias atvejis, nes $\frac{m + 1}{n} + p = \frac{0 + 1}{2} - \frac{1}{2} = 0$ . Čia m=0, n=2, $p = - \frac{1}{2}$ . Keičiame $a + b x^{n} = t^{α} x^{n},$ kur $α$ - trupmenos p vardiklis. Taigi $a + b x^{n} = t^{α} x^{n}$ , čia a=3, b=1; $x^{2} + 3 = t^{2} x^{2}$ ; $3 = t^{2} x^{2} - x^{2}$ ; $x^{2} = \frac{3}{t^{2} - 1} = 3 (t^{2} - 1)^{- 1}$ ; $x = \sqrt{3} \cdot (t^{2} - 1)^{- \frac{1}{2}}$ .

d x = - \frac{\sqrt{3}}{2} \cdot (t^{2} - 1)^{- \frac{3}{2}} \cdot 2 t d t = - \frac{\sqrt{3} t}{\sqrt{(t^{2} - 1)^{3}}} d t .

\int \frac{d x}{\sqrt{x^{2} + 3}} = - \sqrt{3} \int \frac{t d t}{\sqrt{(t^{2} - 1)^{3}} \cdot \sqrt{\frac{3}{t^{2} - 1} + 3}} = - \sqrt{3} \int \frac{t d t}{\sqrt{(t^{2} - 1)^{3}} \cdot \sqrt{3} \sqrt{\frac{1 + t^{2} - 1}{t^{2} - 1}}} =

= - \int \frac{t d t}{\sqrt{(t^{2} - 1)^{3}} \cdot \sqrt{\frac{t^{2}}{t^{2} - 1}}} = - \int \frac{t d t}{\sqrt{(t^{2} - 1)^{3}} \cdot t \sqrt{\frac{1}{t^{2} - 1}}} = - \int \frac{d t}{\sqrt{(t^{2} - 1)^{3}} \cdot \sqrt{\frac{1}{t^{2} - 1}}} =

= - \int \frac{d t}{\sqrt{\frac{(t^{2} - 1)^{3}}{t^{2} - 1}}} = - \int \frac{d t}{\sqrt{(t^{2} - 1)^{2}}} = - \int \frac{d t}{t^{2} - 1} = \int \frac{d t}{1 - t^{2}} .

Iš interentinio integratoriaus:

\int \frac{d t}{1 - t^{2}} = \frac{1}{2} (\ln (t + 1) - \ln (1 - t)) + C = \tanh^{- 1} t + C = artanh t + C .

Kur

t = \sqrt{\frac{x^{2} + 3}{x^{2}}} .

Kad galima būtų skaičiuoti pagal šią formulę t turi būti mažiau už 1 (t<1). Nes kitaip nesiskaičiuoja ln(1-t). Bet mūsų pavyzdyje, kad ir kokias x reikšmes nestatysi į

t = \sqrt{\frac{x^{2} + 3}{x^{2}}},

vistiek t bus daugiau už 1.

Galima taip integruot:

- \int \frac{d t}{t^{2} - 1} = - \int \frac{d t}{(t - 1) (t + 1)};

toliau integruojama kaip racionali funkcija.

\frac{- 1}{(t - 1) (t + 1)} = \frac{A}{t - 1} + \frac{B}{t + 1};

Abi lygties puses padauginame iš (t-1)(t+1). Tada

- 1 = A (t + 1) + B (t - 1),

- 1 = A t + A + B t - B,

- 1 = (A + B) t + A - B;

iš čia turime sistemą:

A+B=0,

A-B=-1.

Tada iš antros lygties A=B-1. Įstačius šia A reikšmę į pirmą lygtį, gauname

B-1+B=0,

2B=1,

B=1/2.

Tada A=-B=-1/2.

Tokiu budu gauname, kad

\frac{- 1}{(t - 1) (t + 1)} = \frac{- 1 / 2}{t - 1} + \frac{1 / 2}{t + 1} .

Integruodami gauname:

- \int \frac{d t}{t^{2} - 1} = - \int \frac{d t}{(t - 1) (t + 1)} = \int (\frac{- 1 / 2}{t - 1} + \frac{1 / 2}{t + 1}) d t = \frac{1}{2} (- \ln (t - 1) + \ln (t + 1)) + C .

Kur

t = \sqrt{\frac{x^{2} + 3}{x^{2}}} .

1. Jei įstatysime x=2, tai gausime,

t = \sqrt{\frac{x^{2} + 3}{x^{2}}} = \sqrt{\frac{2^{2} + 3}{2^{2}}} = \sqrt{\frac{7}{4}} = \sqrt{1.75} = 1.32287565553 .

\frac{1}{2} (- \ln (t - 1) + \ln (t + 1)) = \frac{1}{2} (- \ln (\sqrt{1.75} - 1) + \ln (\sqrt{1.75} + 1)) = 0.986646961 =

=0.98664696104483410110205523811797.

Kai x=1, tai

t = \sqrt{\frac{x^{2} + 3}{x^{2}}} = \sqrt{\frac{1^{2} + 3}{1^{2}}} = \sqrt{\frac{4}{1}} = \sqrt{4} = 2 .

\frac{1}{2} (- \ln (t - 1) + \ln (t + 1)) = \frac{1}{2} (- \ln (2 - 1) + \ln (2 + 1)) = \frac{\ln 3}{2} = 0.549306144334 =

=0.54930614433405484569762261846126.

Bet tokia funkcija integruojama lengviau kitaip (ne per diferencialinius binomus; integruojant keičiant kintamąjį) ir yra jinai integralų lentelėje

\int \frac{d x}{\sqrt{x^{2} \pm a^{2}}} = \ln | x + \sqrt{x^{2} \pm a^{2}} | + C .

2. Tada, kai x=2, tai

\ln | x + \sqrt{x^{2} + 3} | = \ln | 2 + \sqrt{2^{2} + 3} | = \ln | 2 + \sqrt{7} | = 1.5359531 =

=1.5359531053788889467996778565792.

O kai x=1, tai

\ln | x + \sqrt{x^{2} + 3} | = \ln | 1 + \sqrt{1^{2} + 3} | = \ln | 1 + \sqrt{4} | = \ln | 3 | = 1.0986122886681 =

=1.0986122886681096913952452369225.

Tada, kai x kinta nuo 1 iki 2, tai pirmu atveju integruojant gauname:

0.98664696104483410110205523811797 - 0.54930614433405484569762261846126 = 0.43734081671077925540443261965671.

Antru atveju, kai x kinta nuo 1 iki 2 integruojant gautume:

1.5359531053788889467996778565792 - 1.0986122886681096913952452369225 = 0.4373408167107792554044326196567.

Abiais būdais integruojant gavome tą patį atsakymą.

Toks Free Pascal kodas:

  var a:longint; c,d:real;
  begin
  for a:=1 to 100000000  do
  d:=d+0.00000002/sqrt(sqr(a*0.00000002)+3);
  for a:=1 to 100000000  do
  c:=c+0.00000001/sqrt(sqr(a*0.00000001)+3);
  writeln(d);
  writeln(c);
  writeln(d-c);
  readln;
  end.

Duoda rezultatus:

9.8664695905111544E-001

5.4930614394743249E-001

4.3734081510368294E-001

po 4 sekundžių su 4.16 GHz dažniu veikiančiu procesorium (per pirmus du paleidimus duoda šituos rezultatus po 18 sekundžių; bet jeigu iškart exe failą (diferencialiniaibinomai.exe) paleist [kurį sukuria visada Free Pascal programa] iš "C:\FPC\3.2.0\bin\i386-win32", tai rezultatai gaunami po 4 sekundžių ir taip yra su visais Free Pascal skaičiavimais, kad per exe failą greičiau skaičiuoja [iš pirmo karto]). Matome, kad rezultatai tokie patys kaip skaičiuojant/integruojant pirmu atveju.

Toks Free Pascal kodas:

  var a:longint; c,d:real;
  begin
  for a:=1 to 1000000000  do
  d:=d+0.000000002/sqrt(sqr(a*0.000000002)+3);
  writeln(d);
  readln;
  end.

duoda rezultatą "9.8664696084608883E-001" (tai reiškia

9.8664696084608883 \cdot 1 0^{- 1}

) po 20 sekundžių su 4.16 GHz dažniu veikiančiu procesoriumi (per pirmus 2 kartus duodą rezultatą po 34 sekundžių).

Kodas apskaičiuoja plotą, po funkciją

f (x) = \frac{1}{\sqrt{x^{2} + 3}},

apribotą šios funkcijos kreive, ašimi Ox, ašimi Oy ir ašiai Ox statmena tiese taške x=2.

Beje, integruojant antru budu, kai x kinta nuo 0 iki 2, gauname:

\int_{0}^{2} \frac{d x}{\sqrt{x^{2} + 3}} = \ln | x + \sqrt{x^{2} + 3} | |_{0}^{2} = \ln | 2 + \sqrt{2^{2} + 3} | - \ln | 0 + \sqrt{0^{2} + 3} | = \ln | 2 + \sqrt{7} | - \ln | \sqrt{3} | =

= 1.5359531053788889467996778565792 - 0.54930614433405484569762261846126 = 0.98664696104483410110205523811797.

Į anksčiau pirmu budu gautą integralą, įstatę

t = \sqrt{\frac{x^{2} + 3}{x^{2}}} = \frac{\sqrt{x^{2} + 3}}{x},

gauname:

\frac{1}{2} (- \ln (t - 1) + \ln (t + 1)) = \frac{1}{2} (- \ln (\frac{\sqrt{x^{2} + 3}}{x} - 1) + \ln (\frac{\sqrt{x^{2} + 3}}{x} + 1)) = \frac{1}{2} \ln \frac{\frac{\sqrt{x^{2} + 3}}{x} + 1}{\frac{\sqrt{x^{2} + 3}}{x} - 1} =

= \frac{1}{2} \ln \frac{\frac{\sqrt{x^{2} + 3} + x}{x}}{\frac{\sqrt{x^{2} + 3} - x}{x}} = \frac{1}{2} \ln \frac{\sqrt{x^{2} + 3} + x}{\sqrt{x^{2} + 3} - x} = \frac{1}{2} (- \ln (\sqrt{x^{2} + 3} - x) + \ln (\sqrt{x^{2} + 3} + x)) .

Tokiu budu gavome, kad

\int \frac{d x}{\sqrt{x^{2} + 3}} = \frac{1}{2} (- \ln (\sqrt{x^{2} + 3} - x) + \ln (\sqrt{x^{2} + 3} + x)) + C .

O bendru atveju gauname tokį, tikriausiai niekam nematytą, integralą:

\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \frac{1}{2} (- \ln (\sqrt{x^{2} + a^{2}} - x) + \ln (\sqrt{x^{2} + a^{2}} + x)) + C .

Integruodami uždavinio sąlygos integralą nuo 0 iki 2, gauname:

\int_{0}^{2} \frac{d x}{\sqrt{x^{2} + 3}} = \frac{1}{2} (- \ln (\sqrt{x^{2} + 3} - x) + \ln (\sqrt{x^{2} + 3} + x)) |_{0}^{2} =

= \frac{1}{2} (- \ln (\sqrt{2^{2} + 3} - 2) + \ln (\sqrt{2^{2} + 3} + 2)) - \frac{1}{2} (- \ln (\sqrt{0^{2} + 3} - 0) + \ln (\sqrt{0^{2} + 3} + 0)) =

= \frac{1}{2} (- \ln (\sqrt{7} - 2) + \ln (\sqrt{7} + 2)) - \frac{1}{2} (- \ln \sqrt{3} + \ln \sqrt{3}) = \frac{1}{2} (- \ln (\sqrt{7} - 2) + \ln (\sqrt{7} + 2)) =

= \frac{1}{2} (- (- 0.43734081671) + 1.535953105) = \frac{1}{2} \cdot 1.973293922 =

=0.98664696104483410110205523811797.

Toks pat atsakymas, kaip ir integruojant anksčiau.

Dar galima gauti kitokia šio integralo išraišką. Štai taip:

\frac{1}{2} \ln \frac{\sqrt{x^{2} + 3} + x}{\sqrt{x^{2} + 3} - x} = \frac{1}{2} \ln \frac{(\sqrt{x^{2} + 3} + x) (\sqrt{x^{2} + 3} + x)}{(\sqrt{x^{2} + 3} - x) (\sqrt{x^{2} + 3} + x)} = \frac{1}{2} \ln \frac{(\sqrt{x^{2} + 3} + x)^{2}}{(x^{2} + 3) - x^{2}} =

= \frac{1}{2} \ln \frac{(\sqrt{x^{2} + 3} + x)^{2}}{3} = - \frac{1}{2} \ln | 3 | + \frac{1}{2} \ln | (\sqrt{x^{2} + 3} + x)^{2} | = - \frac{1}{2} \ln | 3 | + \ln | \sqrt{x^{2} + 3} + x | .

Tada

\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = - \ln | a | + \ln | \sqrt{x^{2} + a^{2}} + x | + C .

Integruodami iš pirmo būdo ką tik gautą integralą nuo 0 iki x, turėsime:

\int_{0}^{x} \frac{d x}{\sqrt{x^{2} + a^{2}}} = (- \ln | a | + \ln | \sqrt{x^{2} + a^{2}} + x |) |_{0}^{x} = (- \ln | a | + \ln | \sqrt{x^{2} + a^{2}} + x |) - (- \ln | a | + \ln | \sqrt{0^{2} + a^{2}} + 0 |) =

= (- \ln | a | + \ln | \sqrt{x^{2} + a^{2}} + x |) - (- \ln | a | + \ln | \sqrt{a^{2}} |) = - \ln | a | + \ln | \sqrt{x^{2} + a^{2}} + x | .

Antru budu integruodami nuo 0 iki x, turime:

\int_{0}^{x} \frac{d x}{\sqrt{x^{2} + a^{2}}} = \ln | x + \sqrt{x^{2} + a^{2}} | |_{0}^{x} = \ln | x + \sqrt{x^{2} + a^{2}} | - \ln | 0 + \sqrt{0^{2} + a^{2}} | = \ln | x + \sqrt{x^{2} + a^{2}} | - \ln | \sqrt{a^{2}} | = \ln | x + \sqrt{x^{2} + a^{2}} | - \ln | a | .

Gavome tokius pačius integralus ir įstačius vietoj x ir a bet kokias reikšmes, abiais būdais gausime tokias pat išraiškas ir atsakymus.

$\int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} d x = \int x^{- 1 / 2} (1 + x^{1 / 4})^{1 / 3} d x;$

m = - \frac{1}{2}, n = \frac{1}{4}, p = \frac{1}{3}, \frac{m + 1}{n} = \frac{- 1 / 2 + 1}{1 / 4} = \frac{1 / 2}{1 / 4} = 2

(II atvejis).

Keičiame

a + b x^{n} = t^{α},

kur

α

- trupmenos p vardiklis. Tada pakeitimas yra

1 + x^{1 / 4} = t^{3},

x^{1 / 4} = t^{3} - 1,

x = (t^{3} - 1)^{4},

d x = 4 (t^{3} - 1)^{3} \cdot 3 t^{2} d t = 12 t^{2} (t^{3} - 1)^{3} d t .

\sqrt[3]{1 + x^{1 / 4}} = t .

\int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} d x = \int x^{- 1 / 2} (1 + x^{1 / 4})^{1 / 3} d x = \int (t^{3} - 1)^{- 2} (1 + (t^{3} - 1))^{1 / 3} 12 t^{2} (t^{3} - 1)^{3} d t =

= \int (1 + (t^{3} - 1))^{1 / 3} 12 t^{2} (t^{3} - 1) d t = 12 \int (t^{3})^{1 / 3} t^{2} (t^{3} - 1) d t = 12 \int t^{3} (t^{3} - 1) d t = 12 \int (t^{6} - t^{3}) d t =

= 12 (\frac{t^{7}}{7} - \frac{t^{4}}{4}) + C = \frac{3}{7} t^{4} (4 t^{3} - 7) + C = \frac{3}{7} (\sqrt[3]{1 + x^{1 / 4}})^{4} (4 (1 + x^{1 / 4}) - 7) + C = \frac{3}{7} \sqrt[3]{1 + x^{1 / 4}} (1 + x^{1 / 4}) (4 (1 + x^{1 / 4}) - 7) + C .

Apskaičiuosime integralą

I = \int \frac{d x}{x^{2} \sqrt{a + b x^{2}}} = \int x^{- 2} (a + b x^{2})^{- \frac{1}{2}} d x .

Šiame pavyzdyje

m = - 2, n = 2, p = - \frac{1}{2},

todėl

\frac{m + 1}{n} + p = \frac{- 2 + 1}{2} - \frac{1}{2} = \frac{- 1}{2} - \frac{1}{2} = - 1

(III atvejis).

Tada

a + b x^{n} = t^{α} x^{n},

kur

α

- trupmenos p vardiklis;

a + b x^{2} = t^{2} x^{2} .

Ir pasinaudojame keitiniu

t = \sqrt{\frac{a}{x^{2}} + b}; t^{2} x^{2} - b x^{2} = a, (t^{2} - b) x^{2} = a, x = \frac{\sqrt{a}}{\sqrt{t^{2} - b}}; d x = - \frac{1}{2} \frac{\sqrt{a} 2 t d t}{(t^{2} - b)^{3 / 2}} = \frac{- \sqrt{a} t d t}{\sqrt{(t^{2} - b)^{3}}} .

Tada gauname

I = \int x^{- 2} (a + b x^{2})^{- \frac{1}{2}} d x = \int \frac{t^{2} - b}{a} (a + b \frac{a}{t^{2} - b})^{- \frac{1}{2}} \frac{- \sqrt{a} t d t}{\sqrt{(t^{2} - b)^{3}}} = \int \frac{t^{2} - b}{a} (\frac{a (t^{2} - b) + a b}{t^{2} - b})^{- \frac{1}{2}} \frac{- \sqrt{a} t d t}{\sqrt{(t^{2} - b)^{3}}} =

= \int \frac{t^{2} - b}{a} (\frac{t^{2}}{t^{2} - b})^{- \frac{1}{2}} \frac{- \sqrt{a} t d t}{\sqrt{(t^{2} - b)^{3}}} = \int \frac{t^{2} - b}{a} \frac{\sqrt{t^{2} - b}}{t} \frac{- \sqrt{a} t d t}{\sqrt{(t^{2} - b)^{3}}} = - \int \frac{d t}{\sqrt{a}} =

= - \frac{t}{\sqrt{a}} + C = - \frac{\sqrt{\frac{a}{x^{2}} + b}}{\sqrt{a}} + C .

Apskaičiuosime integralą $I = \int x^{5} (1 - x^{2})^{- \frac{1}{2}} d x .$ Šiuo atveju $m = 5, n = 2, p = - \frac{1}{2},$ todėl $\frac{m + 1}{n} = 3$ (II atvejis). Pasinaudoję pakeitimais

t = \sqrt{1 - x^{2}}, t^{2} = 1 - x^{2}, x = \sqrt{1 - t^{2}}, d x = \frac{- 2 t d t}{2 \sqrt{1 - t^{2}}} = \frac{- t d t}{\sqrt{1 - t^{2}}},

gauname

I = \int x^{5} (1 - x^{2})^{- \frac{1}{2}} d x = \int (\sqrt{1 - t^{2}})^{5} (1 - [1 - t^{2}])^{- \frac{1}{2}} \frac{- t d t}{\sqrt{1 - t^{2}}} = \int \sqrt{1 - t^{2}} (1 - t^{2})^{2} (t^{2})^{- \frac{1}{2}} \frac{- t d t}{\sqrt{1 - t^{2}}} =

= \int \sqrt{1 - t^{2}} (1 - t^{2})^{2} \frac{1}{t} \frac{- t d t}{\sqrt{1 - t^{2}}} = - \int (1 - t^{2})^{2} d t = - \int (1 - 2 t^{2} + t^{4}) d t = - \int d t + 2 \int t^{2} d t - t^{4} d t =

= - t + \frac{2}{3} t^{3} - \frac{t^{5}}{5} + C = - \sqrt{1 - x^{2}} + \frac{2}{3} \sqrt{(1 - x^{2})^{3}} - \frac{\sqrt{(1 - x^{2})^{5}}}{5} + C .

Nuorodos

Oilerio keitiniai
https://en.wikipedia.org/wiki/Euler_substitution Oilerio keitiniai

Matematika/Iracionaliųjų funkcijų integravimas: Skirtumas tarp puslapio versijų

Dabartinė 18:13, 30 balandžio 2023 versija

Turinys

Papildomai

Oilerio keitiniai

Pavyzdžiai

Diferencialinių binomų integravimas

Nuorodos

Naršymo meniu

Matematika/Iracionaliųjų funkcijų integravimas: Skirtumas tarp puslapio versijų

Dabartinė 18:13, 30 balandžio 2023 versija

Papildomai

Oilerio keitiniai

Pavyzdžiai

Diferencialinių binomų integravimas

Nuorodos

Naršymo meniu

Paieška