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

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

The correct answer is Option (1) → $\frac{1}{4} [\log |x^2 - 1| - \log |x^2 + 1|] + C$

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

Put $x^2 = t \Rightarrow 2x \, dx = dt \Rightarrow x \, dx = \frac{1}{2} dt$

$∴I = \frac{1}{2} \int \frac{dt}{t^2 - 1} = \frac{1}{2} \cdot \frac{1}{2 \cdot 1} \log \left| \frac{t - 1}{t + 1} \right| + C \quad \left[ ∵\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C \right]$

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