$\cfrac{1}{{x - {x^3}}}$

Let $I = \int {\cfrac{{dx}}{{x - {x^3}}}}$ and

Let $\cfrac{1}{{x - {x^3}}} = \cfrac{1}{{x\left( {x + 1} \right)\left( {1 - x} \right)}} = \cfrac{A}{x} + \cfrac{B}{{1 + x}} + \cfrac{C}{{1 - x}}$

$\Rightarrow$ $1 = A\left( {1 + x} \right)\left( {1 - x} \right) + Bx\left( {1 - x} \right) + Cx\left( {1 + x} \right)$

….(i)
Putting $x = 0$ in (i),

we get
$1 = A\left( {1 + 0} \right)\left( {1 - 0} \right)$ $\Rightarrow$ $A = 1$

Putting $x = - 1$ in (i),

we get
$1 = B\left( { - 1} \right)\left( {1 + 1} \right)$ $\Rightarrow$ $B = - \cfrac{1}{2}$

Putting $x = 1$ in (i),

we get
$1 = C\left( 1 \right)\left( {1 + 1} \right)$ $\Rightarrow$ $C = \cfrac{1}{2}$

$\therefore$ $\cfrac{1}{{x - {x^3}}} = \cfrac{1}{x} - \cfrac{1}{{2\left( {1 + x} \right)}} + \cfrac{1}{{2\left( {1 - x} \right)}}$

$\Rightarrow$ $I = \int {\cfrac{1}{{x - {x^3}}}} dx = \int {\cfrac{1}{x}dx} - \cfrac{1}{2}\int {\cfrac{1}{{1 + x}}} dx + \cfrac{1}{2}\int {\cfrac{1}{{1 - x}}} dx$

$= \log \left| x \right| - \cfrac{1}{2}\log \left| {1 + x} \right| - \cfrac{1}{2}\log \left| {1 - x} \right| + C$

$= \log \left| x \right| - \cfrac{1}{2}\log \left| {1 - {x^2}} \right| + C = \cfrac{1}{2}\log \left| {\cfrac{{{x^2}}}{{1 - {x^2}}}} \right| + C$