Let f (x) = ∫ x (1 + x) ^ 2 dx(x ≥ 0) . Then f(3) - f(1) is equal...

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

Let $x = \tan^{2}\theta$

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

$f\left(x\right) = ∫ \dfrac{tan\theta}{{\left(1 + \tan^{2}\theta\right)}^{2}} . 2tan\theta{sec}^{2}\thetad\theta$

$f\left(x\right) = ∫ \dfrac{tan\theta}{\sec^{4}\theta} . 2tan\theta{sec}^{2}\thetad\theta$

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

$f\left(x\right) = ∫ 2\sin^{2}\thetad\theta$

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

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

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

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

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