Question

Let $f\left(x\right)=\int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})}dx$ and $f\left(0\right)=0$. Then $f\left(1\right)$ is equal to

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

Answer 1

Answer: (B)

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

Let
$f\left(x\right)=I=\int \frac{x^2}{(x^2+1)(1+\sqrt{x^2+1})}dx=\int \frac{\sqrt{x^2+1}-1}{x^2+1}dx=\int (\frac{1}{\sqrt{x^2+1}}-\frac{1}{x^2+1})dx=I_1+I_2$
Where
$I_1=\int \frac{1}{\sqrt{x^2+1}}dx$
Substituting $x=\tan t\Rightarrow dx={\sec }^{2}tdt$
$I_1=\int \sec tdt=\int \frac{{\sec }^{2}t+\tan t\sec t}{\tan t+\sec t}dt$
Again substituting $u=\tan t+\sec t\Rightarrow du=({\sec }^{2}t+\tan t\sec t)dt$
$I_1=\int \frac{1}{u}=\log u+c=\log (\sqrt{x^2+1}+x)+c$
And $I_2=\int \frac{1}{x^2+1}dx=-{\tan }^{-1}x+c$
Hence
$f\left(x\right)=I=\log (\sqrt{x^2+1}+x)-{\tan }^{-1}x+Cf\left(0\right)=0\Rightarrow \log 1+0+C=0\Rightarrow C=0f\left(1\right)=\log (\sqrt{2}+1)-\frac{\pi }{4}$