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 $\left(\sqrt{1+x^{2}}\right)=\mathrm{t}$
$\frac{x}{\sqrt{1+x^{2}}} d x=d t$
$I=\int \frac{x^{3}}{\sqrt{1+x^{2}}} d x=\int x^{2} d t=\int\left(t^{2}-1\right) d t$
$I=\frac{t^{3}}{3}-t\left[\right.$ put $\left.(t)=\sqrt{1+x^{2}}\right]$
$I=\frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{3}-\sqrt{1+x^{2}}+C$
$\left[a=\frac{1}{3}\right] ;[b=-1]$