Question

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

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

Answer 1

Answer: (D)

$f(x)=\underset{1}{\overset{3}{\int }}\frac{\sqrt{x}dx}{(1+x{)}^2}=\underset{1}{\overset{\sqrt{3}}{\int }}\frac{t\cdot 2tdt}{{\left(1+t^2\right)}^2}($ put $\sqrt{x}=t)$
$={\left(-\frac{t}{1+t^2}\right)}_1^{\sqrt{3}}+{\left({\tan }^{-1}t\right)}_1^{\sqrt{3}}$ [Appling by parts]
$=-\left(\frac{\sqrt{3}}{4}-\frac{1}{2}\right)+\frac{\pi }{3}-\frac{\pi }{4}$
$=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{\pi }{12}$