Integrals — Class 12 Maths Solution

Let $I = \int {\frac{x}{{{x^4} - 1}}} dx$
Let's put ${x^2} = t \Rightarrow 2xdx = dt \Rightarrow xdx = \frac{1}{2}dt$

therefore,$I = \frac{1}{2}\int {\frac{{dt}}{{{t^2} - 1}}} = \frac{1}{2} \cdot \frac{1}{2}\log \left| {\frac{{t - 1}}{{t + 1}}} \right| + C$

$= \frac{1}{4}\left[ {\log \left| {{x^2} - 1} \right| - \log \left| {{x^2} + 1} \right|} \right] + C$