Question

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

Answer 1

$\frac{1}{x-x^3}$=$\frac{1}{x-x^2}$=$\frac{1}{x(1-x)(1+x)}$

Let $\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x}.........(i)$

⇒1=A-Ax2+Bx+Bx2+Cx-Cx2

Equating the coefficients of x2,x, and the constant term, we obtain

-A+B-C=0

B+C=0

A=1

In 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)

⇒∫$\frac{1}{x}$(1-x)(1+x)dx=∫$\frac{1}{x}$dx+$\frac{1}{2}$∫$\frac{1}{1-x}$dx-$\frac{1}{2}$∫$\frac{1}{1+x}$dx

log|x|-log|(1-x)$^{\frac{1}{2}}$|-log|(1+x)$^{\frac{1}{2}}$|

=log|$\frac{x^2}{1-x^2}{^{\frac{1}{2}}}$|+C

=$\frac{1}{2}$log|$\frac{x^2}{1-x^2}$|+C