Evaluate $\displaystyle \int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} \, dx$

$\frac{\pi^2}{32}$

The correct answer is Option (3) → $\frac{\pi^2}{32}$

Let $t = \tan^{-1} x$, then $dt = \displaystyle \frac{1}{1 + x^2} \, dx$. The new limits are, when $x = 0, t = 0$ and when $x = 1, t = \displaystyle \frac{\pi}{4}$. Thus, as $x$ varies from $0$ to $1$, $t$ varies from $0$ to $\displaystyle \frac{\pi}{4}$.

Therefore $\int\limits_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} \, dx = \int\limits_{0}^{\frac{\pi}{4}} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{\frac{\pi}{4}}= \frac{1}{2} \left[ \frac{\pi^2}{16} - 0 \right] = \frac{\pi^2}{32}$