Evaluate $\int \frac{x^2}{1 - x^4} dx$

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

The correct answer is Option (3) → $\frac{1}{4} \log \left| \frac{1 + x}{1 - x} \right| - \frac{1}{2} \tan^{-1} x + C$

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

$= \int \frac{\left( \frac{1}{2} + \frac{x^2}{2} - \frac{1}{2} + \frac{x^2}{2} \right)}{(1 - x^2)(1 + x^2)} dx \quad [∵a^2 - b^2 = (a+b)(a-b)] \text{}$

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

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

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

$= \frac{1}{4} \log \left| \frac{1 + x}{1 - x} \right| - \frac{1}{2} \tan^{-1} x + C \quad [∵C = C_1 + C_2] \text{}$