Question

Let $f\left(x\right)=\int \frac{\sqrt{x}}{{\left(1+x\right)}^2}dx,x\ge 0$. Then $f\left(3\right)-f\left(1\right)$ is equal to?

• A. $-\frac{\mathrm{\pi }}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$
• B. $\frac{\mathrm{\pi }}{6}+\frac{1}{2}-\frac{\sqrt{3}}{4}$
• C. $-\frac{\mathrm{\pi }}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$
• D. $\frac{\mathrm{\pi }}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

Answer 1

The correct option is D

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

Explanation for the correct solution

Given that: $f\left(x\right)=\int \frac{\sqrt{x}}{{\left(1+x\right)}^2}dx,x\ge 0$

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

differentiating with respect to $t$,

$dx=2.\tan t.se{c}^{2}tdt$

$$\begin{gathered} f\left(x\right)=\int \frac{\sqrt{x}}{{\left(1+x\right)}^2}dx \\ =\int \frac{\sqrt{{\tan }^{2}t}}{{(1+{\tan }^{2}t)}^2}2.\tan t.se{c}^{2}tdt \\ =\int \frac{\tan t}{se{c}^{4}t}2.\tan t.se{c}^{2}tdt \\ =2\int {\sin }^{2}t.dt \\ =\int \left(1-\cos 2t\right)dt \\ =t-\frac{\sin 2t}{2} \end{gathered}$$

Now, $$\begin{gathered} \mathrm{when}x=3,t=\frac{\mathrm{\pi }}{3}, \\ \mathrm{when}x=1,t=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Therefore,

$$\begin{gathered} f\left(3\right)-f\left(1\right)=\left[\left(\frac{\mathrm{\pi }}{3}\right)-\left(\frac{\mathrm{\pi }}{4}\right)\right]-\frac{1}{2}\left[\sin \left(\frac{2\mathrm{\pi }}{3}\right)-\sin \left(\frac{\mathrm{\pi }}{2}\right)\right] \\ =\frac{\mathrm{\pi }}{12}-\frac{1}{2}\left[\frac{\sqrt{3}}{2}-1\right] \\ =\frac{\mathrm{\pi }}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4} \end{gathered}$$

Hence, the correct answer is option (D).