Question

Let $f\left(x\right)=\int \frac{x^{2}dx}{(1+x^2)(1+\sqrt{1+x^2})}$ and $f\left(0\right)=0$. Then the value of $f\left(1\right)$ will be

• A. $\log (1+\sqrt{2})$
• B. $\log (1+\sqrt{2})-\frac{\pi }{4}$
• C. $\log (1+\sqrt{2})+\frac{\pi }{2}$
• D. None of these

Answer 1

Answer: (B)

$\log (1+\sqrt{2})-\frac{\pi }{4}$

$f\left(x\right)=\int \frac{x^{2}dx}{(1+x^2)(1+\sqrt{1+x^2})}$
Let $x=\tan \theta$
$\Rightarrow dx={\sec }^2\theta d\theta =(1+x^2)d\theta$
$\therefore f\left(x\right)=\int \frac{x^{2}dx}{(1+x^2)(1+\sqrt{1+x^2})}$
$=\int \frac{{\tan }^2\theta {\sec }^2\theta d\theta }{{\sec }^2\theta (1+\sec \theta )}$
$=\int \frac{{\tan }^2\theta d\theta }{1+\sec \theta }$
$=\int \frac{{\sin }^2\theta d\theta }{\cos \theta (1+\cos \theta )}$
$=\int \frac{1-{\cos }^2\theta d\theta }{\cos \theta (1+\cos \theta )}$
$=\int \frac{(1-\cos \theta )d\theta }{\cos \theta }$
$=\int \sec \theta d\theta -\int d\theta$
$=\log |\sec \theta +\tan \theta |-\theta +C$
$=\log (x+\sqrt{1+x^2})-{\tan }^{-1}x+C$
Given $f\left(0\right)=0$
$\Rightarrow 0=\log 1-0+C$
$\Rightarrow C=0$
$f\left(x\right)=\log (x+\sqrt{1+x^2})-{\tan }^{-1}x$
$f\left(1\right)=\log (1+\sqrt{1+1})-{\tan }^{-1}\left(1\right)$
$\Rightarrow f\left(1\right)=\log (1+\sqrt{2})-\frac{\pi }{4}$