$\int \frac{1}{\sqrt{x^2+2}} d\left(x^2+1\right)$ is equal to

$2 \sqrt{x^2+2}+C$ $\sqrt{x^2+2}+C$ $\frac{1}{\left(x^2+2\right)^{3 / 2}}+C$ none of these

$2 \sqrt{x^2+2}+C$

We have, $I=\int \frac{1}{\sqrt{x^2+2}} d\left(x^2+1\right)$ $\Rightarrow I=\int \frac{1}{\sqrt{x^2+2}} d\left(x^2+2\right)$                $\left[∵ d\left(x^2+1\right)=d\left(x^2+2\right)\right]$ $\Rightarrow I=2 \sqrt{x^2+2}+C$