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

$\int \frac{\sqrt{x}}{(1+x{)}^2}dx$
Put $\sqrt{x}=\tan t$
$\Rightarrow dx=2\left(\tan t\right)\left({\sec }^{2}t\right)dt$
$\therefore \int \frac{\left(\tan t\right)\left(2\tan t\right)\left({\sec }^{2}t\right)dt}{{\sec }^{4}t}$
$=\int 2{\sin }^{2}tdt$
$=\int (1-\cos 2t)dt$
$=t-\frac{\sin 2t}{2}+C$
At $x=3\Rightarrow t=\frac{\pi }{3}$
At $x=1\Rightarrow t=\frac{\pi }{4}$
Now, $f\left(3\right)-f\left(1\right)=(\frac{\pi }{3}-\frac{\sqrt{3}}{4})-(\frac{\pi }{4}-\frac{1}{2})$
$\therefore f\left(3\right)-f\left(1\right)=\frac{\pi }{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$