Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (b)/Solution 3

We apply integration by parts with $u={\frac {x^{2}}{2}}$ and $dv=2x(1+x^{2})^{\frac {1}{2}}\,dx$. Then $du=x\,dx$ and $v={\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}$. Thus

${\begin{aligned}\int x^{3}{\sqrt {1+x^{2}}}dx&=\int {\frac {x^{2}}{2}}\cdot 2x(1+x^{2})^{\frac {1}{2}}dx\\&={\frac {x^{2}}{2}}{\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}-\int x{\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}dx\\&={\frac {x^{2}}{2}}{\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}-{\frac {1}{3}}{\frac {2}{5}}(1+x^{2})^{\frac {5}{2}}+C\\&={\frac {x^{2}}{3}}(1+x^{2})^{\frac {3}{2}}-{\frac {2}{15}}(1+x^{2})^{\frac {5}{2}}+C\end{aligned}}$

Note: The fastest way to confirm that this is the same answer as we get in the solutions above, check that they both simplify to ${\sqrt {1+x^{2}}}\left(-{\frac {2}{15}}+{\frac {1}{15}}x^{2}+{\frac {1}{5}}x^{4}\right)+C$.