Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (b)/Solution 1

For the change of variable $u=1+x^{2}$, we have

$\displaystyle du=2x\,dx.$

So

$x^{3}{\sqrt {1+x^{2}}}dx=x^{2}{\sqrt {1+x^{2}}}\,x\,dx=x^{2}{\sqrt {u}}\,{\frac {du}{2}}$

To write the remaining $x^{2}$ in terms of $u$, we note that since $u=1+x^{2}$ we have $x^{2}=u-1$. Thus

${\begin{aligned}\int x^{3}{\sqrt {1+x^{2}}}dx&=\int (u-1){\sqrt {u}}\,{\frac {du}{2}}\\&=\int (u-1)u^{1/2}\,{\frac {du}{2}}\\&={\frac {1}{2}}\int (u^{3/2}-u^{1/2})du\\&={\frac {1}{2}}\left({\frac {2}{5}}u^{5/2}-{\frac {2}{3}}u^{3/2}\right)+C\\&={\frac {1}{2}}\left({\frac {2}{5}}(1+x^{2})^{5/2}-{\frac {2}{3}}(1+x^{2})^{3/2}\right)+C\\&={\frac {1}{5}}(1+x^{2})^{5/2}-{\frac {1}{3}}(1+x^{2})^{3/2}+C.\end{aligned}}$