Integrals — Class 12 Maths Solution

Let $I = \int {\cfrac{{{x^2}dx}}{{1 - {x^6}}}} = \int {\cfrac{{{x^2}}}{{1 - {{\left( {{x^3}} \right)}^2}}}dx}$

Put ${x^3} = t$ $\Rightarrow$ $3{x^2}dx = dt$

$\therefore$ $I = \cfrac{1}{3}\int {\cfrac{{dt}}{{1 - {t^2}}}} = \cfrac{1}{3}.\cfrac{1}{2}\log \left| {\cfrac{{1 + t}}{{1 - t}}} \right| + C$

$= \cfrac{1}{6}\log \left| {\cfrac{{1 + t}}{{1 - t}}} \right| + C = \cfrac{1}{6}\log \left| {\cfrac{{1 + {x^3}}}{{1 - {x^3}}}} \right| + C$