Find: $\int \frac{1}{x(x^2 - 1)} dx$

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

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

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

Let $x^2 = t ⇒2x dx = dt$

$= \frac{1}{2} \int \frac{dt}{t(t - 1)}$

$= \frac{1}{2} \int \left[ \frac{1}{t - 1} - \frac{1}{t} \right] dt$

$= \frac{1}{2} [\log |t - 1| - \log |t|] + C$

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