Question

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

• A. $\log (1+\sqrt{2})$
• B. $\log (1+\sqrt{2})+\frac{\pi }{4}$
• C. $\log (1+\sqrt{2})-\frac{\pi }{4}$
• D. $\log (\sqrt{2}-1)$

Answer 1

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

To find $\int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})}dx$,

Substitute $x=\tan \theta$ and $dx={\sec }^2\theta d\theta$

$=\int \frac{{\tan }^2\theta }{{\sec }^2\theta (1+\sec \theta )}{\sec }^2\theta d\theta =\int \frac{{\sin }^2\theta }{\cos \theta (1+\cos \theta )}d\theta =\int \frac{4{\sin }^2\frac{\theta }{2}.{\cos }^2\frac{\theta }{2}}{\cos \theta .2{\cos }^2\frac{\theta }{2}}d\theta =\int \frac{2{\sin }^2\frac{\theta }{2}}{\cos \theta }d\theta =\int \frac{1-\cos \theta }{\cos \theta }d\theta =\int (\sec \theta -1)d\theta =\log (\sqrt{1+x^2}+x)-{\tan }^{-1}x+c$

Given that $f\left(0\right)=0$. Substituting in the expression for $f\left(x\right)$ we get $c=0$.

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