Mark the correct alternative in each of the following:
Question:

Mark the correct alternative in each of the following:

$\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=a\left(1+x^{2}\right)^{3 / 2}+b \sqrt{1+x^{2}}+C$, then

A. $a=\frac{1}{3}, b=1$

B. $a=-\frac{1}{3}, b=1$

C. $a=-\frac{1}{3}, b=-1$

D. $a=\frac{1}{3}, b=-1$

Solution:

let $\left(\sqrt{1+x^{2}}\right)=\mathrm{t}$

$\frac{x}{\sqrt{1+x^{2}}} d x=d t$

$I=\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=\int x^{2} d t=\int\left(t^{2}-1\right) d t$

$I=\frac{t^{3}}{3}-t\left[\operatorname{put}(t)=\sqrt{1+x^{2}}\right]$

$I=\frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{3}-\sqrt{1+x^{2}}+C$

$\left[\mathrm{a}=\frac{1}{3}\right] ;[\mathrm{b}=-1]$