Integration #1

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

Therefore, $a+b+c = 1+2+4 = \boxed{7}$ .