Question

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

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

Answer 1

The correct option is D $\frac{\pi }{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

Given:
$f\left(x\right)=\int \frac{\sqrt{x}}{(1+x{)}^2}dx$
Put $x={\tan }^2\theta$
$\Rightarrow dx=2\tan \theta {\sec }^2\theta d\theta$
$\therefore f\left(\theta \right)=\int \frac{2{\tan }^2\theta .{\sec }^2\theta }{{\sec }^4\theta }d\theta$
$\Rightarrow f\left(\theta \right)=\int 2{\sin }^2\theta d\theta$
$\Rightarrow f\left(\theta \right)=\theta -\frac{\sin 2\theta }{2}+C$
$\Rightarrow f\left(\theta \right)=\theta -\frac{1}{2}\times \frac{2\tan \theta }{1+{\tan }^2\theta }+C$
$\Rightarrow f\left(\theta \right)=\theta -\frac{\tan \theta }{1+{\tan }^2\theta }+C$
$\therefore f\left(x\right)={\tan }^{-1}\sqrt{x}-\frac{\sqrt{x}}{1+x}+C$
Now,
$f\left(3\right)-f\left(1\right)={\tan }^{-1}\left(\sqrt{3}\right)-\frac{\sqrt{3}}{1+3}-{\tan }^{-1}\left(1\right)+\frac{1}{2}$
$\therefore f\left(3\right)-f\left(1\right)=\frac{\pi }{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$