Integration of $f(x)=\sqrt{1+x^2}$ with respect to $x^2$, is

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

$\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C$

Let $I =\int f(x) d\left(x^2\right)=\int \sqrt{1+x^2} d\left(x^2\right)$ $\Rightarrow I =\int \sqrt{1+x^2} d\left(1+x^2\right) ~~~~~\left[∵ d\left(x^2\right)=d\left(1+x^2\right)\right]$ $\Rightarrow I=\frac{2}{3}\left(1+x^2\right)^{3 / 2}+C$