$$
\begin{aligned}
& \mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0 \\
& \mathrm{f}(3)-\mathrm{f}(1)=\int_1^3 \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x=\mathrm{I}(\text { say })
\end{aligned}
$$

Put $\sqrt{x}=\tan \theta \Rightarrow \mathrm{d} x=2 \tan \theta \sec ^2 \theta \mathrm{d} \theta$
When $x=1, \theta=\frac{\pi}{4}$ and when $x=3, \theta=\frac{\pi}{3}$
$$
\begin{aligned}
\therefore \quad I & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 \tan ^2 \theta \sec ^2 \theta}{\left(1+\tan ^2 \theta\right)^2} d \theta \\
& =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 \tan ^2 \theta}{1+\tan ^2 \theta} d \theta \\
& =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 \sin ^2 \theta+\cos ^2 \theta}{} d \theta \\
& =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}(1-\cos 2 \theta) \mathrm{d} \theta \\
& =\left[\theta-\frac{\sin 2 \theta}{2}\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} \\
& =\left(\frac{\pi}{3}-\frac{\pi}{4}\right)-\frac{1}{2}\left(\frac{\sqrt{3}}{2}-1\right) \\
& =\frac{\pi}{12}-\frac{\sqrt{3}}{4}+\frac{1}{2}
\end{aligned}
$$