Question

Integrate the function $\frac{1}{x-x^3}$

Answer 1

$\frac{1}{x-x^3}=\frac{1}{x(1-x^2)}=\frac{1}{x(1-x)(1+x)}$
Let $\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x}$ ........ (1)
$\Rightarrow 1=A(1-x^2)+Bx(1+x)+Cx(1-x)$
$\Rightarrow 1=A-A{x}^2+Bx+B{x}^2+Cx-C{x}^2$
Equating the coefficients of $x^2,x$, and constant term, we obtain
$A+BC=0$
$B+C=0$
$A=1$
On solving these equations, we obtain
$A=1,B=\frac{1}{2}$, and $C=-\frac{1}{2}$
From equation (1), we obtain
$\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}$
$\Rightarrow \int \frac{1}{x(1-x)(1+x)}dx=\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{1-x}dx-\frac{1}{2}\int \frac{1}{1+x}dx$
$=\log |x|-\frac{1}{2}\log |(1-x)|-\frac{1}{2}\log |(1+x)|$
$=\log |x|-\log |(1-x{)}^{\frac{1}{2}}|-\log |(1+x{)}^{\frac{1}{2}}|$
$=\log \left|\frac{x}{(1-x{)}^{\frac{1}{2}}(1+x{)}^{\frac{1}{2}}}\right|+C$
$=\log \left|{\left(\frac{x^2}{1-x^2}\right)}^{\frac{1}{2}}\right|+C$
$=\frac{1}{2}\log \left|\frac{x^2}{1-x^2}\right|+C$