The value of $\int\limits_{-1}^1 \frac{\log \left(x+\sqrt{1+x^2}\right)}{x+\log \left(x+\sqrt{1+x^2}\right)} f(x) d x-\int\limits_{-1}^1 \frac{\log \left(x+\sqrt{1+x^2}\right)}{x+\log \left(x+\sqrt{1+x^2}\right)} f(-x) d x$, is
Answer & explanation
Correct answer: option 1
Let $g(x)=\frac{\log \left(x+\sqrt{1+x^2}\right)}{x+\log \left(x+\sqrt{1+x^2}\right)}$ and let $I$ be the value of the given expression.
Then,
$I=\int\limits_{-1}^1 g(x) f(x) d x-\int\limits_{-1}^1 g(x) f(-x) d x$
$\Rightarrow I =\int\limits_{-1}^1 g(x)\{f(x)-f(-x)\} d x$
$\Rightarrow I=\int\limits_{-1}^1 g(x) h(x) d x$, where $h(x)=f(x)-f(-x)$
We observe that $g(x)$ and $h(x)$ are even and odd functions respectively. Therefore, $g(x) h(x)$ is an odd function and hence,
$I=\int\limits_{-1}^1 g(x) h(x) d x=0$