Integrals — Class 12 Maths Solution

Let $I = \int {\frac{{{x^2}}}{{1 - {x^4}}}} dx$

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

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

$= \frac{1}{2}\int {\frac{1}{{1 - {x^2}}}} dx - \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} dx = \frac{1}{2} \cdot \frac{1}{2}\log \left| {\frac{{1 + x}}{{1 - x}}} \right| + {C_1} - \frac{1}{2}{\tan ^{ - 1}}x + {C_2}$