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 $f\left(1\right)$ is

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

Answer 1

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

$f\left(x\right)=\int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})}dx$
$=\int \frac{\sqrt{1+x^2}-1}{1+x^2}dx=\int \frac{1}{\sqrt{1+x^2}}dx-\int \frac{1}{1+x^2}dx$
$=\ln |x+\sqrt{1+x^2}|-{tan}^{-1}x+C$
From $f\left(0\right)=C=0$
Substituting $C=0$ we get
$f\left(1\right)=\ln |1+\sqrt{1+1}|-{tan}^{-1}1+0=\ln |1+\sqrt{2}|-\frac{\pi }{4}$