7.1 Integration By Parts/30

$\int _{0}^{1}{\frac {r^{3}}{\sqrt {4+r^{2}}}}\cdot dr$

${\begin{aligned}u&=4+r^{2}\\[2ex]r^{2}&=u-4\\[2ex]2r\cdot dr&=du\\[2ex]\end{aligned}}$

$\int _{0}^{1}{\frac {r^{3}}{\sqrt {4+r^{2}}}}\cdot dr~~~=~~~\int _{0}^{1}{\frac {r}{2{\sqrt {u}}}}\cdot du~~~=~~~\int _{0}^{1}{\frac {u-4}{2{\sqrt {u}}}}\cdot du~~~=~~~{\frac {}{}}{\frac {1}{2}}\int _{0}^{1}\left({\frac {u}{\sqrt {u}}}-{\frac {4}{\sqrt {u}}}\right)\cdot du~~~=~~~$

${\frac {1}{2}}\left[\left({\frac {u^{{\frac {1}{2}}+1}}{{\frac {1}{2}}+1}}\right)-\left({\frac {u^{-{\frac {1}{2}}+1}}{-{\frac {1}{2}}+1}}\right)\right]~~~=~~~{\frac {1}{2}}\left[\left({\frac {u^{\frac {3}{2}}}{\frac {3}{2}}}\right)-4\left({\frac {u^{\frac {1}{2}}}{\frac {1}{2}}}\right)\right]~~~=~~~{\frac {u^{\frac {3}{2}}}{3}}-4u^{\frac {1}{2}}$

Now, we need to substitute u back

${\frac {\left(r^{2}+4\right)^{\frac {3}{2}}}{3}}-4\left(r^{2}+4\right)^{\frac {1}{2}}+C$

$\int _{0}^{1}{\frac {r^{3}}{\sqrt {4+r^{2}}}}\cdot dr~~~=~~~\left[{\frac {\left(r^{2}+4\right)^{\frac {3}{2}}}{3}}-4\left(r^{2}+4\right)^{\frac {1}{2}}\right]$

$\left[{\frac {\left(1^{2}+4\right)^{\frac {3}{2}}}{3}}-4\left(1^{2}+4\right)^{\frac {1}{2}}\right]-\left[{\frac {\left(0^{2}+4\right)^{\frac {3}{2}}}{3}}-4\left(0^{2}+4\right)^{\frac {1}{2}}\right]$

<math> \frac{5^{\frac{3}{2}}}{3}-4\left ( 5 \right )^{\frac{1}{2}}-\left ( \frac{\left ( 4 \right )^{\frac{3}{2}}}{3} + 8\right ) <math>