Задача №1598

\[ \frac{x^2}{1-x^4} =\frac{-x^2}{(x-1)(x+1)\left(x^2+1\right)} =\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}=\\ =\frac{A(x+1)\left(x^2+1\right)+B(x-1)\left(x^2+1\right)+(Cx+D)(x-1)(x+1)}{(x-1)(x+1)\left(x^2+1\right)} \]

\[ -x^2=A(x+1)\left(x^2+1\right)+B(x-1)\left(x^2+1\right)+(Cx+D)(x-1)(x+1) \]

\[ \begin{aligned} & x=1;\;-1=4A;\;A=-\frac{1}{4}.\\ & x=-1;\;-1=-4B;\;B=\frac{1}{4}.\\ & x=0;\;0=A-B-D=-D-\frac{1}{2};\;D=-\frac{1}{2}.\\ & x=2;\;-4=15A+5B+6C+3D=6C-4;\;C=0. \end{aligned} \]

\[ \int\frac{x^2dx}{1-x^4} =\int\left(-\frac{1}{4}\cdot\frac{1}{x-1}+\frac{1}{4}\cdot\frac{1}{x+1}-\frac{1}{2}\cdot\frac{1}{x^2+1}\right)dx =-\frac{1}{4}\ln|x-1|+\frac{1}{4}\ln|x+1|-\frac{1}{2}\arctg{x}+C \]