∫ 0 1 r 3 4 + r 2 ⋅ d r {\displaystyle \int _{0}^{1}{\frac {r^{3}}{\sqrt {4+r^{2}}}}\cdot dr}
u = 4 + r 2 r 2 = u − 4 2 r ⋅ d r = d u {\displaystyle {\begin{aligned}u&=4+r^{2}\\[2ex]r^{2}&=u-4\\[2ex]2r\cdot dr&=du\\[2ex]\end{aligned}}}
∫ 0 1 r 3 4 + r 2 ⋅ d r = ∫ 0 1 r 2 u ⋅ d r {\displaystyle \int _{0}^{1}{\frac {r^{3}}{\sqrt {4+r^{2}}}}\cdot dr~~~=~~~\int _{0}^{1}{\frac {r}{2{\sqrt {u}}}}\cdot dr}
![{\displaystyle {\begin{aligned}u&=4+r^{2}\\[2ex]r^{2}&=u-4\\[2ex]2r\cdot dr&=du\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/52ab8b5de0b47d5a125b84a3af2c6619c3c4367b)
