Question

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

• A. $\frac{1}{4}\log\left|\frac{x+1}{x-1}\right| - \frac{1}{2}\tan^{-1}x + c$
• B. $\frac{1}{4}\log\left|\frac{x-1}{x+1}\right| + \frac{1}{2}\tan^{-1}x + c$
• C. $\frac{1}{4}\log\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\tan^{-1}x + c$
• D. $\frac{1}{4}\log\left|\frac{x+1}{x-1}\right| + \frac{1}{2}\tan^{-1}x + c$

Hint:

Integration using Partial Fractions.

• Decompose rational expressions into simpler fractions.
• Integrate each part using standard results.
• Watch out for even powers in denominators — often lead to $\tan^-1$.

Answer 1

The Correct Option is B

Split the integrand into partial fractions: \[ \frac{x^2}{(x^2-1)(x^2+1)} = \frac{1}{2(x^2-1)} + \frac{1}{2(x^2+1)} \] Further, decompose: \[ \frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) \] So: \[ \int \frac{x^2}{(x^2-1)(x^2+1)} dx = \frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| + \frac{1}{2}\tan^{-1}x + c \]