7.1 Integration By Parts/30

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

${\begin{aligned}u&=4+r^{2}\\[2ex]du&={\frac {1}{\sqrt {1-x^{2}}}}\;dx\\[2ex]\end{aligned}}$

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