$\int \frac{1}{\sqrt{x^2+2}} d\left(x^2+1\right)$ is equal to
Answer & explanation
Correct answer: option 1
We have,
$I=\int \frac{1}{\sqrt{x^2+2}} d\left(x^2+1\right)$
$\Rightarrow I=\int \frac{1}{\sqrt{x^2+2}} d\left(x^2+2\right)$ $\left[∵ d\left(x^2+1\right)=d\left(x^2+2\right)\right]$
$\Rightarrow I=2 \sqrt{x^2+2}+C$