Question

Let $f(x)=\int \frac{\sqrt{x}}{(1+x)^{2}} d x(x \geq 0) .$ 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)

$f(x)=\int\limits_{1}^{3} \frac{\sqrt{x} d x}{(1+x)^{2}}=\int\limits_{1}^{\sqrt{3}} \frac{t \cdot 2 t d t}{\left(1+t^{2}\right)^{2}} \quad($ put $\sqrt{x}=t)$
$=\left(-\frac{t}{1+t^{2}}\right)_{1}^{\sqrt{3}}+\left(\tan ^{-1} t\right)_{1}^{\sqrt{3}} \quad$ [Appling by parts]
$=-\left(\frac{\sqrt{3}}{4}-\frac{1}{2}\right)+\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{\pi}{12}$