Let f x = ∫ x 1 + x 2 d x     x ≥ 0 . Then f 3 - f 1 is equal to :

$f\left(x\right)=\int \frac{\sqrt{x}}{{\left(1+x\right)}^2}dx$

Let $x={\tan }^2\theta$

$dx=2\tan \theta {\sec }^2\theta d\theta$

$f\left(x\right)=\int \frac{\tan \theta }{{\left(1+{\tan }^2\theta \right)}^2}.2\tan \theta {\sec }^2\theta d\theta$

$f\left(x\right)=\int \frac{\tan \theta }{{\sec }^4\theta }.2\tan \theta {\sec }^2\theta d\theta$

$f\left(x\right)=\int 2{\tan }^2\theta .{\cos }^2\theta d\theta$

$f\left(x\right)=\int 2{\sin }^2\theta d\theta$

$f\left(x\right)=\int \left(1-\cos 2\theta \right) d\theta$

$f\left(x\right)=\theta -\frac{\sin 2\theta }{2}+C=\theta -\frac{\tan \theta }{1+{\tan }^2\theta }+C$

$f\left(x\right)={\tan }^{-1}\sqrt{x}-\frac{\sqrt{x}}{1+x}+C$

Now, $$$f\left(3\right)-f\left(1\right)={\tan }^{-}\sqrt{3}-\frac{\sqrt{3}}{1+3}-{\tan }^{-1}\sqrt{1}+\frac{1}{1+1}$

$=\frac{\pi }{3}-\frac{\pi }{4}+\frac{1}{2}-\frac{\sqrt{3}}{4}=\frac{\pi }{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$