Let $f(x)=\int \frac{x^2}{\left(1+x^2\right)\left\{1+\sqrt{1+x^2}\right\}} d x$ and $f(0)=0$. Then, $f(1)$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$f(x)=\int \frac{x^2}{\left(x^2+1\right)\left\{1+\sqrt{1+x^2}\right\}} d x$
$\Rightarrow f(x)=\int \frac{\left(\sqrt{1+x^2}-1\right)}{\left(x^2+1\right)} d x=\int \frac{1}{\sqrt{x^2+1}} d x-\int \frac{1}{x^2+1} d x$
$\Rightarrow f(x)=\log _e\left(x+\sqrt{x^2+1}\right)-\tan ^{-1} x+C$
Now, $f(0)=0 \Rightarrow 0=\log _e 1-\tan ^{-1} 0+C \Rightarrow C=0$
∴ $f(x)=\log _e\left(x+\sqrt{x^2+1}\right)-\tan ^{-1} x$
$\Rightarrow f(1)=\log _e(\sqrt{2}+1)-\frac{\pi}{4}$