Piern

$\begin{aligned} f(x) & = \int \frac {32x^2}{(x^2+1)^2} dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = 32 \int \frac {\tan^2 \theta \sec^2 \theta}{\sec^6 \theta} d \theta \\ & = 32 \int \frac {\tan^2 \theta}{\sec^4 \theta} d \theta \\ & = 32 \int \sin^2 \theta \cos^2 \theta \ d \theta \\ & = 8 \int \sin^2 2\theta \ d \theta \\ & = 4 \int (1-\cos 4 \theta) \ d\theta \\ & = 4 \theta - \sin 4 \theta + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ & = 4 \theta - 2 \sin 2\theta \cos 2\theta + C \\ & = 4 \theta - 2 \sin \theta \cos \theta (1 - 2\sin^2 \theta) + C & \small \color{#3D99F6} \text{Note that } \tan \theta = x \implies \\ \implies f(x) & = 4 \tan^{-1} x - \frac {2x(1-x^2)}{1+x^2} + C & \small \color{#3D99F6} \sin \theta = \frac x{\sqrt{1+x^2}}, \ \cos \theta = \frac 1{\sqrt{1+x^2}} \\ f(0) & = 4(0) - 0 + C = 0 & \small \color{#3D99F6} \implies C = 0 \\ f(1) & = 4 \tan^{-1} 1 - 0 + 0 \\ & = \pi \end{aligned}$

$\implies a = \boxed{1}$