Integrate the function w.r.t. $x$: $\frac{\sin (\tan^{-1} x)}{1 + x^2}$

$-\cos (\tan^{-1} x) + C$

The correct answer is Option (4) → $-\cos (\tan^{-1} x) + C$

Derivative of $\tan^{-1} x = \frac{1}{1 + x^2}$. Thus, we use the substitution

$\tan^{-1} x = t$ so that $\frac{dx}{1 + x^2} = dt$.

Therefore, $\int \frac{\sin (\tan^{-1} x)}{1 + x^2} \, dx = \int \sin t \, dt = -\cos t + C = -\cos (\tan^{-1} x) + C$