Question

If $f\left(x\right)=\int \frac{x^{2}dx}{(1+x^2)(1+\sqrt{1+x^2})}$ and $f\left(0\right)=0$, then the value of $f\left(1\right)$ is

• 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

The correct option is B $\log (1+\sqrt{2})-\frac{\pi }{4}$

$f\left(x\right)=\int \frac{x^{2}dx}{(1+x^2)(1+\sqrt{1+x^2})}$

Substitute $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta =(1+x^2)d\theta$

$\therefore f\left(x\right)=\int \frac{x^{2}dx}{(1+x^2)(1+\sqrt{1+x^2})}=\int \frac{{\tan }^2\theta {\sec }^2\theta d\theta }{{\sec }^2\theta (1+\sec \theta )}$

$=\int \frac{{\tan }^2\theta d\theta }{1+\sec \theta }=\int \frac{{\sin }^2\theta d\theta }{\cos \theta (1+\cos \theta )}=\int \frac{1-{\cos }^2\theta d\theta }{\cos \theta (1+\cos \theta )}$

$=\int \frac{(1-\cos \theta )}{\cos \theta }d\theta =\int \sec \theta d\theta -\int d\theta =\log (x+\sqrt{1+x^2})-{\tan }^{-1}x+cf\left(0\right)=\log (0+\sqrt{1+0})-{\tan }^{-1}0+c\Rightarrow 0=\log 1-0+c\Rightarrow c=0$

$\therefore f\left(1\right)=\log (1+\sqrt{1+1^2})-{\tan }^{-1}1=\log (1+\sqrt{2})-\frac{\pi }{4}$