Question

Integrate the following functions w.r.t. x.

$\int \frac{1}{x-x^3}dx$

Answer 1

$\int \frac{1}{x-x^3}dx=\int \frac{1}{x(1-x^2)}dx=\int \frac{1}{x(1-x)(1+x)}...\left(i\right)Let\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{(1+x)}\Rightarrow 1=A(1-x)(1+x)+B\left(x\right)(1+x)+C\left(x\right)(1-x)\Rightarrow 1=A(1-x^2)+B(x+x^2)+C(x-x^2)\Rightarrow 1=x^2(B-A-C)+x(B+C)+A\text{On equating the coefficients of}{x}^2\text{, x and constant term on both sides, we get}-A+B-C=0,B+C=0andA=1\text{On solving these equations, we get A=1, B=}\frac{1}{2}andC=-\frac{1}{2}FromEq.\left(i\right),weget\int \frac{1}{x(1-x^2)}dx=\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{1-x}-\frac{1}{2}\int \frac{1}{1+x}dx=log|x|-\frac{1}{2}log|1-x|-\frac{1}{2}log|1+x|+C=log|x|-\frac{1}{2}log\{|1+x||1-x|\}+C=\frac{1}{2}|x^2|-\frac{1}{2}log|1-x^2|+C=\frac{1}{2}log|\frac{x^2}{1-x^2}|+C$