Question

Решить неопределённый интеграл методом рациональных функций : 1)интеграл x^2-x+14/(x-2)(x-4)^3 dx2)интеграл dx/x^4-x^2-2

Answer 1

$1)\; \; \frac{x^2-x+14}{(x-2)(x-4)^3}=\frac{A}{x-2}+\frac{B}{x-4}+\frac{C}{(x-4)^2}+\frac{D}{(x-4)^3}\; ,\\\\x^2-x+14=A(x-4)^3+B(x-2)(x-4)^2+C(x-2)(x-4)+D(x-2)\; ,\\\\x=2:\; \; A=\frac{4-2+14}{-8}=-2\; ,\\\\x=4:\; \; D=\frac{16-4+14}{2}=13\; ,\\\\x^3\, |\; A+B=0\; \; \to \; \; B=-A=2\\\\x^2\, |\; -12A-10B+C=1\; \; \to \; \; C=1-24+20=-3\\\\\\\int \frac{x^2-x+14}{(x-2)(x-4)^3}=\int \frac{-2\, dx}{x-2}+\int \frac{2\, dx}{x-4}+\int \frac{-3\, dx}{(x-4)^2}+\int \frac{13\, dx}{(x-4)^3}=$

$-2ln|x-2|+2ln|x-4|-3\cdot \frac{(x-4)^{-1}}{-1}+13\cdot \frac{(x-4)^{-2}}{-2}+C=\\\\=-2ln|x-2|+2ln|x-4|+\frac{3}{x-4}-\frac{13}{2(x-4)^2}+C=\\\\=ln\Big (\frac{x-4}{x-2}\Big )^2+\frac{3}{x-4}-\frac{13}{2(x-4)^2}+C\; ;$

$2)\; \; \frac{1}{x^4-x^2-2}=\frac{1}{(x^2-2)(x^2+1)}=\frac{1}{(x-\sqrt2)(x+\sqrt2)(x^2+1)}=\\\\=\frac{A}{x-\sqrt2}+\frac{B}{x+\sqrt2}+\frac{Cx+D}{x^2+1}\; ,\\\\1=A(x+\sqrt2)(x^2+1)+B(x-\sqrt2)(x^2+1)+(Cx+D)(x^2-2)\; ,\\\\x=\sqrt2:\; \; A=\frac{1}{6\sqrt2}\; ,\; \; \; x=-\srqt2:\; \; B=-\frac{1}{6\sqrt2}\\\\x^3:\; A+B+C=0\; \; \to \; \; C=-(A+B)=0\\\\x^0:\; A\sqrt2-B\sqrt2-2D=1\; \; \to \; \; 2D=\frac{1}{6}+\frac{1}{6}-1=-\frac{2}{3}\; ,\; D=-\frac{1}{3}$

$\int \frac{dx}{x^4-x62-2}=\frac{1}{6\sqrt2}\int \frac{dx}{x-\sqrt2}-\frac{1}{6\sqrt2}\int \frac{dx}{x+\sqrt2}-\frac{1}{3}\int \frac{dx}{x^2+1}=\\\\=\frac{1}{6\sqrt2}\cdot ln|x-\sqrt2|-\frac{1}{6\sqrt2}\cdot ln|x+\sqrt2|-\frac{1}{3}arctgx+C=\\\\=\frac{1}{6\sqrt2}\cdot ln\Big |\frac{x-\sqrt2}{x+\sqrt2}\Big |-\frac{1}{3}arctgx+C\; .$