Question

Solve:

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

Answer 1

$I=\int \frac{1}{x(1-x^2)}dx=\int \frac{1}{x(1-x)(1+x)}dx$

using parctial fraction, we have

$I=\int \frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}$

on comparing numerator, we have

In $A(1-x)(1+x)+Bx(1+x)+C(1-x)x=1$

put $x=1,B=1/2putx=0,A=1$

put $x=-1,C=-1/2$

$\Rightarrow I=\int (\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)})dx$

$=lnx-\frac{1}{2}ln(1-x)+\frac{1}{2}ln(1+x)$

$=ln\left(\frac{x\sqrt{1+x}}{\sqrt{1-x}}\right)$