Integration of $f(x)=\sqrt{1+x^2}$ with respect to $x^2$, is
Answer & explanation
Correct answer: option 2
Let
$I =\int f(x) d\left(x^2\right)=\int \sqrt{1+x^2} d\left(x^2\right)$
$\Rightarrow I =\int \sqrt{1+x^2} d\left(1+x^2\right) ~~~~~\left[∵ d\left(x^2\right)=d\left(1+x^2\right)\right]$
$\Rightarrow I=\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C$