Matematika/Racionaliųjų funkcijų integravimas

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Šis straipsnis yra apie racionaliųjų funkcijų integravimą.

[redaguoti] Integravimas funkcijų turinčių kvadratinį trinarį

$\int{dx\over ax^2+bx+c}=\int{4a\;dx\over 4a^2 x^2+4abx+b^2+4ac-b^2}=4a\int{dx\over (2ax+b)^2+(4ac-b^2)}=$

$={4a\over 2a}\int{du\over u^2+s}={2\over \sqrt{s}} \arctan{u\over \sqrt{s}}+C={2\over\sqrt{4ac-b^2}}\arctan{2ax+b\over\sqrt{4ac-b^2}}+C,$ kur $u = 2 a x + b;$ $d u = 2 a d x;$ $4 a c - b 2 = s;$ s>0.

$\int{Mx+N\over ax^2+bx+c}dx=4a\int{Mx+N\over (2ax+b)^2+(4ac-b^2)}dx={4a\over 2a}\int{M\frac{u-b}{2a}+N\over u^2+s}du=$ $={1\over a}\int{Mu-Mb+2aN\over u^2+s}du={M\over a}\int{u\;du\over u^2+s}+{2aN-Mb\over a}\int{du\over u^2+s}=$ $={M\over 2a}\int{d(u^2+s)\over u^2+s}+{2aN-Mb\over a}\int{du\over u^2+s}={M\over 2a}\ln|u^2+s|+{2aN-Mb\over a \sqrt{s}}\arctan{u\over \sqrt{s}}+C=$ $={M\over 2a}\ln|(2ax+b)^2+4ac-b^2|+{2aN-Mb\over a \sqrt{4ac-b^2}}\arctan{2ax+b\over \sqrt{4ac-b^2}}+C=$ $={M\over 2a}\ln|4a(ax^2+bx+c)|+{2aN-Mb\over a \sqrt{4ac-b^2}}\arctan{2ax+b\over \sqrt{4ac-b^2}}+C,$ kur $u = 2 a x + b;$ $x={u-b\over 2a};$ $d u = 2 a d x;$ $4 a c - b 2 = s;$ $s > 0;$ $d (u 2 + s) = 2 d u .$

$\int{Ax+B\over ax^2+bx+c}dx=\int{{A\over 2a}(2ax+b)+(B-{Ab\over 2a})\over ax^2+bx+c}dx=$

$={A\over 2a}\int{2ax+b\over ax^2+bx+c}dx+(B-{Ab\over 2a})\int{dx\over ax^2+bx+c}.$

$\int{dx\over ax^2+bx+c}={1\over a}\int{dx\over (x+{b\over 2a})^2+({c\over a}-{b^2\over 4a^2})}.$

$ax^2+bx+c=a(x^2+{b\over a}x+{c\over a})=a[x^2+2{b\over 2a}x+({b\over 2a})^2+{c\over a}-({b\over 2a})^2]=a[(x+{b\over 2a})^2+({c\over a}-{b^2\over 4a^2})].$

$\int{Ax+B\over \sqrt{ax^2+bx+c}}dx=\int{{A\over 2a}(2ax+b)+(B-{Ab\over 2a})\over\sqrt{ax^2+bc+c}}dx=$

$={A\over 2a}\int{2ax+b\over\sqrt{ax^2+bx+c}}dx+(B-{Ab\over 2a})\int{dx\over\sqrt{ax^2+bx+c}}.$ Pritaikę pirmajam iš gautų integralų keitinį $a x 2 + b x + c = t,$ $(2 a x + b) d x = d t,$ gauname: $\int{(2ax+b)dx\over\sqrt{ax^2+bx+c}}=\int{dt\over \sqrt{t}}=2\sqrt{t}+C=2\sqrt{ax^2+bx+c}+C.$

[redaguoti] Pavyzdžiai

$\int{6x+5\over x^2+4x+9}dx=\int\frac{6x+5}{(x+2)^2+5}dx=\int{6(t-2)+5\over t^2+5}dt=\int{6t-7\over t^2+5}dt=3\int{d(t^2+5)\over t^2+5}-$

$-7\int{dt\over t^2+5}=3\ln(t^2+5)-\frac{7}{\sqrt{5}}\arctan{t\over \sqrt{5}}+C=3\ln(x^2+4x+9)-{7\over \sqrt{5}}\arctan{x+2\over \sqrt{5}}+C,$

kur $x + 2 = t;$ $x = t - 2;$ $d x = d t .$

$\int{7x-2\over 3x^2-5x+4}dx=\int{84x-24\over 36x^2-60x+48}dx=\int{84x-24\over (6x-5)^2+23}dx=\frac{1}{6}\int{14u+70-24\over u^2+23}du=$

$={7\over 3}\int{u\;du\over u^2+23}+{23\over 3}\int{du\over u^2+23}=\frac{7}{6}\ln|u^2+23|+\frac{23}{3\sqrt{23}}\arctan{u\over \sqrt{23}}du+C=$ $={7\over 6}\ln|36x^2-60x+48|+{\sqrt{23}\over 3}\arctan {6x-5\over\sqrt{23}}+C={7\over 6}\ln|3x^2-5x+4|+{\sqrt{23}\over 3}\arctan {6x-5\over\sqrt{23}}+C_1,$

kur $u = 6 x - 5;$ $d u = 6 d x;$ $x={u+5\over 6}.$

$\int{5x-3\over x^2+6x-40}dx=\int{5x-3\over (x+3)^2-49}dx=\int{5u-15-3\over u^2-49}du=5\int{u\; du\over u^2-49}-$

$-18\int{du\over u^2-49}dx={5\over 2}\ln|u^2-49|-{18\over 14}\ln|{u-7\over u+7}|+C={5\over 2}\ln|x^2+6x-40|-{9\over 7}\ln|{x-4\over x+10}|+C,$

kur $u = x + 3;$ $d u = d x;$ $x = u - 3.$ Šį uždavinį galima išspresti ir naudojantis aukščiau pateikta formule: $\int{5x-3\over x^2+6x-40}dx=\int{4(5x-3)dx\over 4x^2+24x+36-160-36}=\int{4(5x-3)dx\over(2x+6)^2-196}={4\over 2}\int{5{u-6\over 2}-3\over u^2-196}du=$ $={2\over 2}\int{5u-30-6\over u^2-196}du=5\int{u\; du\over u^2-196}-36\int{du\over u^2-196}={5\over 2}\int{d(u^2-196)\over u^2-196}-{36\over 28}\ln|{u-14\over u+14}|+C=$ $={5\over 2}\ln|u^2-196|-{9\over 7}\ln|{2x-8\over 2x+20}|+C_1={5\over 2}\ln|4(x^2+6x-40)|-{9\over 7}\ln|{x-4\over x+10}|+C_1,$ kur $2 x + 6 = u;$ $x={u-6\over 2};$ $d u = 2 d x;$ $d (u 2 - 196) = 2 u d u .$ Abiejų būdų atsakymai gali skirtis konstanta.

$\int{3x+4\over \sqrt{7-x^2+6x}}dx=\int{3x+4\over\sqrt{16-(x-3)^2}}dx=\int{3u+9+4\over\sqrt{16-u^2}}du=\int{3u\;du\over\sqrt{16-u^2}}+\int{13du\over\sqrt{16-u^2}}=$

$=-3\int\sqrt{16-u^2}+13\arcsin{u\over 4}+C=-3\sqrt{7-x^2+6x}+13\arcsin{x-3\over 4}+C,$ kur $u = x - 3;$ $d u = d x;$ $x = u + 3.$

$\int{(x+3)dx\over x^2-2x-5}=\int{{1\over 2}(2x-2)+(3+{1\over 2}2)\over x^2-2x-5}dx={1\over 2}\int{(2x-2)dx\over x^2-2x-5}+4\int{dx\over x^2-2x-5}=$

$={1\over 2}\int{d(x^2-2x-5)\over x^2-2x-5}+4\int{dx\over (x-1)^2-6}={1\over 2}\ln|x^2-2x-5|+4{1\over\sqrt{6}}\ln|{\sqrt{6}-(x-1)\over\sqrt{6}+(x-1)}|+C.$

$\int{dx\over 2x^2+8x+20}={1\over 2}\int{dx\over x^2+4x+10}={1\over 2}\int{dx\over 1[(x+{4\over 2\cdot 1})^2+({10\over 1}-{4^2\over 4\cdot 1^2})]}=$

$={1\over 2}\int{dx\over (x^2+4x+4)+(10-4)}={1\over 2}\int{dx\over (x+2)^2+6}={1\over 2}\int{dt\over t^2+6}={1\over 2}{1\over\sqrt{6}}\arctan{t\over\sqrt{6}}+C=$ $={1\over 2\sqrt{6}}\arctan{x+2\over\sqrt{6}}+C,$ kur $x + 2 = t,$ $d x = d t .$

$\int{5x+3\over\sqrt{x^2+4x+10}}dx=\int{{5\over 2}(2x+4)+(3-10)\over\sqrt{x^2+4x+10}}dx=$

$={5\over 2}\int{2x+4\over\sqrt{x^2+4x+10}}dx-7\int{dx\over \sqrt{(x+2)^2+6}}=$ $={5\over 2}\int{d(x^2+4x+10)\over\sqrt{x^2+4x+10}}-7\ln|x+2+\sqrt{(x+2)^2+6}|+C=$ $=5\sqrt{x^2+4x+10}-7\ln|x+2+\sqrt{x^2+4x+10}|+C,$ kur $d (x 2 + 4 x + 10) = (2 x + 4) d x;$ $d (x + 2) = d x .$

[redaguoti] Racionalių trupmenų išskaidymas elementariosiomis

$\int\frac{x^5+x^4-8}{x^3-4x}dx=\int[x^2+x+4+\frac{4x^2+16x-8}{x(x+2)(x-2)}]dx=\frac{x^3}{3}+\frac{x^2}{2}+4x+4\int\frac{x^2+4x-2}{x(x+2)(x-2)}dx.$

$\frac{x^2+4x-2}{x(x+2)(x-2)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}.$

$x 2 + 4 x - 2 = A (x + 2)(x - 2) + B x (x - 2) + C x (x + 2) = A (x 2 - 4) + B (x 2 - 2 x) + C (x 2 + 2 x).$ Iš čia sudarome sistemą:

A + B + C = 1,

- 2 B + 2 C = 4,

- 4 A = - 2.

Iš sistemos randame: $A=\frac{1}{2};\;B=-\frac{3}{4};\;C=\frac{5}{4}.$ Vadinasi,

$4\int\frac{x^2+4x-2}{x(x+2)(x-2)}dx=4(\frac{1}{2}\int\frac{dx}{x}-\frac{3}{4}\int\frac{dx}{x+2}+\frac{5}{4}\int\frac{dx}{x-2})=$ $=4(\frac{1}{2}\ln|x|-\frac{3}{4}\ln|x+2|+\frac{5}{4}\ln|x-2|)+C=\ln|\frac{x^2(x-2)^5}{(x+2)^3}|+C.$ Irašę šią reikšmę į pačią pirmąją lygybę, gauname: $\int\frac{x^5+x^4-8}{x^3-4x}dx=\frac{x^3}{3}+\frac{x^2}{2}+4x+\ln|\frac{x^2(x-2)^5}{(x+2)^3}|+C.$

$\int\frac{xdx}{x^2-6x+12}=\int\frac{xdx}{(x-3)^2+3}.$ Taikydami keitinį $x - 3 = u,$ gauname:

$\int\frac{(u+3)du}{u^2+3}=\int\frac{u\;du}{u^2+3}+3\int \frac{du}{u^2+3}=\frac{1}{2}\int\frac{d(u^2+3)}{u^2+3}+3\int \frac{du}{u^2+(\sqrt{3})^2}=$ $=\frac{1}{2}\ln(u^2+3)+\frac{3}{\sqrt{3}}\arctan \frac{u}{\sqrt{3}}+C=\frac{1}{2}\ln(x^2-6x+12)+\sqrt{3}\arctan\frac{x-3}{\sqrt{3}}+C.$

Šį rezultatą buvo galima gauti iš karto remiantis $\int\frac{Mx+N}{x^2+px+q}dx=\frac{M}{2}\ln|x^2+px+q|+\frac{2N-Mp}{\sqrt{4q-p^2}}\arctan\frac{2x+p}{\sqrt{4q-p^2}}+C$ lygybe. Mūsų atveju $M = 1,$ $N = 0,$ $p = - 6$ ir $q = 12.$ Todėl $\int\frac{xdx}{x^2-6x+12}=\frac{1}{2}\ln(x^2-6x+12)+\frac{-1\cdot(-6)}{\sqrt{48-36}}\arctan\frac{2x-6}{\sqrt{48-36}}+C=$ $=\frac{1}{2}\ln(x^2-6x+12)+\sqrt{3}\arctan\frac{x-3}{\sqrt{3}}+C.$

$\int\frac{15x^2-4x-81}{x^3-13x+12}dx=\int\frac{15x^2-4x-81}{(x-3)(x+4)(x-1)}dx=\int[\frac{A}{x-3}+\frac{B}{x+4}+\frac{D}{x-1}]dx.$

$15 x 2 - 4 x - 81 = A (x + 4)(x - 1) + B (x - 3)(x - 1) + D (x - 3)(x + 4) = A (x 2 + 3 x - 4) + B (x 2 - 4 x + 3) + D (x 2 + x - 12).$ Sulyginę koeficientus prie vienodų x laipsnių, gauname tiesinių lygčių sistemą

A + B + D = 15,

3 A - 4 B + D = - 4,

- 4 A + 3 B - 12 D = - 81.

Iš sistemos randame: $A = 3;$ $B = 5;$ $D = 7.$ Vadinasi $\int\frac{15x^2-4x-81}{x^3-13x+12}dx=3\int\frac{dx}{x-3}+5\int\frac{dx}{x+4}+7\int\frac{dx}{x-1}=$

= 3ln | x - 3 | + 5ln | x + 4 | + 7ln | x - 1 | + C = ln | (x - 3) 3 (x + 4) 5 (x - 1) 7 | + C .

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

$\frac{x+2}{x(x-2)(x+1)}=\frac{A}{x}+\frac{B}{x-2}+\frac{D}{x+1};$ $x + 2 = A (x - 2)(x + 1) + B x (x + 1) + D x (x - 2) = A (x 2 - x - 2) + B (x 2 + x) + D (x 2 - 2 x);$

A + B + D = 0;

- A + B - 2 D = 1;

- 2 A = 2;

$A = - 1;$ $B=\frac{2}{3};$ $D=\frac{1}{3}.$ $\int\frac{x^4-3x^2-3x-2}{x^3-x^2-2x}dx=$ $=\int(x+1)dx+\int\frac{dx}{x}-\frac{2}{3}\int\frac{dx}{x-2}-\frac{1}{3}\int\frac{dx}{x+1}=\frac{x^2}{2}+x+\ln |x|-\frac{2}{3}\ln|x-2|-\frac{1}{3}\ln|x+1|+C.$

$\int\frac{2x^2-3x+3}{x^3-2x^2+x}dx=\int[\frac{A}{x}+\frac{B}{(x-1)^2}+\frac{D}{x-1}]dx.$

$2 x 2 - 3 x + 3 = A (x - 1) 2 + B x + D x (x - 1) = A (x 2 - 2 x + 1) + B x + D (x 2 - x).$

A + D = 2,

- 2 A + B - D = - 3,

A = 3.

$A = 3;$ $B = 2;$ $D = - 1.$ $\int\frac{2x^2-3x+3}{x^3-2x^2+x}dx=3\int\frac{dx}{x}+2\int\frac{dx}{(x-1)^2}-\int\frac{dx}{x-1}=3\ln|x|-\frac{2}{x-1}-\ln|x-1|+C.$

$\int{2x-1\over x^2-5x+6}dx=\int{2x-1\over (x-3)(x-2)}dx=\int({A\over x-3}+{B\over x-2})dx,$

kur $2 x - 1 = A (x - 2) + B (x - 3) = (A + B) x - 2 A - 3 B .$ Sulyginam koeficientus vienodu laipsnių ir turime sistemą:

A + B = 2;

- 2 A - 3 B = - 1.

Iš kur $A = 5;$ $B = - 3.$ Tuomet $\int{2x-1\over x^2-5x+6}dx=\int({5\over x-3}-{3\over x-2})dx=5\int{d(x-3)\over x-3}-3\int{d(x-2)\over x-2}=$ $=5\ln|x-3|-3\ln|x-2|+C=\ln|{(x-3)^5\over (x-2)^3}|+C.$

$\int{9x^3-30x^2+28x-88\over (x^2-6x+8)(x^2+4)}dx=\int{9x^3-30x^2+28x-88\over (x-2)(x-4)(x^2+4)}dx=\int({A\over x-2}+{B\over x-4}+{Cx+D\over x^2+4})dx;$

9 x 3 - 30 x 2 + 28 x - 88 = A (x - 4)(x 2 + 4) + B (x - 2)(x 2 + 4) + (C x + D)(x 2 - 6 x + 8) =

= (A + B + C) x 3 + ( - 4 A - 2 B - 6 C + D) x 2 + (4 A + 4 B + 8 C - 6 D) x + ( - 16 A - 8 B + 8 D);

Palyginam koeficientus prie vienodų laipsnių x.

x 3

|

A + B + C = 9,

x 2

|

- 4 A - 2 B - 6 C + D = - 30,

x 1

|

4 A + 4 B + 8 C - 6 D = 28,

x 0

|

- 16 A - 8 B + 8 D = - 88,

Išsprendę sistemą , randame: $A = 5;$ $B = 3;$ $C = 1;$ $D = 2.$ $\int{9x^3-30x^2+28x-88\over (x^2-6x+8)(x^2+4)}dx=\int({5\over x-2}+{3\over x-4}+{x+2\over x^2+4})dx=5\int{d(x-2)\over x-2}+3\int{d(x-4)\over x-4}+$ $+{1\over 2}\int{d(x^2+4)\over x^2+4}+\int{2\;dx\over x^2+4}=5\ln|x-2|+3\ln|x-4|+{1\over 2}\ln|x^2+4|+{2\over 2}\arctan{x\over 2}+C=$ $=\ln|(x-2)^5(x-4)^3\sqrt{x^2+4}|+\arctan{x\over 2}+C,$ kur $d (x 2 + 4) = 2 x d x .$

$\int{(x^2+2)dx\over (x+1)^3(x-2)}=\int({A\over(x+1)^3}+{A_1\over (x+1)^2}+{A_2\over x+1}+{B\over x-2})dx.$

$x + 2 = A (x - 2) + A 1 (x + 1)(x - 2) + A 2 (x + 1) 2 (x - 2) + B (x + 1) 3 = (A 2 + B) x 3 + (A 1 + 3 B) x 2 + (A - A 1 - 3 A 2 + 3 B) x + ( - 2 A - 2 A 1 - 2 A 2 + B).$

x 3

|

0 = A 2 + B,

x 2

|

1 = A 1 + 3 B,

x 1

|

0 = A - A 1 - 3 A 2 + 3 B,

x 0

|

2 = - 2 A - 2 A 1 - 2 A 2 + B .

Išsprendę sistema, randame: $A=-1;\;A_1={1\over 3};\;A_2=-{2\over 9};\;B={2\over 9}.$ $\int{(x^2+2)dx\over (x+1)^3(x-2)}=\int(-{1\over(x+1)^3}+{1\over 3(x+1)^2}-{2\over 9(x+1)}+{2\over 9(x-2)})dx=$ $={1\over 2(x+1)^2}-{1\over 3(x+1)}-{2\over 9}\ln|x+1|+{2\over 9}\ln|x-2|+C={1-2x\over 6(x+1)^2}+\ln|({x-2\over x+1})^{2\over 9}|+C.$