A calculus problem by Akshat Sharda

$\begin{aligned} I & = \int_0^1 \frac {x\tan^{-1}x}{(1+x^2)^\frac 32} \ dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d \theta \\ & = \int_0^\frac \pi 4 \frac {\theta \tan \theta \sec^2 \theta }{(1+\tan^2 \theta)^\frac 32} \ d \theta & \small \color{#3D99F6} \text{Note that } \sec^2 \theta = 1+ \tan^2 \theta \\ & = \int_0^\frac \pi 4 \frac {\theta \tan \theta}{\sec \theta} \ d \theta & \small \color{#3D99F6} \text{Note that } \sec \theta = \frac 1{\cos \theta} \\ & = \int_0^\frac \pi 4 \theta \sin \theta \ d \theta & \small \color{#3D99F6} \text{By integration by parts} \\ & = - \theta \cos \theta \bigg|_0^\frac \pi 4 + \int_0^\frac \pi 4 \cos \theta \ d \theta \\ & = - \frac \pi 4 \cdot \frac 1{\sqrt 2} + \sin \theta \bigg|_0^\frac \pi 4 \\ & = - \frac {\pi}{4 \sqrt 2} + \frac 1{\sqrt 2} - 0 \\ & = \frac 1{\sqrt 2} - \frac {\pi}{4 \sqrt 2} \end{aligned}$

$\implies n = \boxed{2}$