∫ 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 u = ∫ 0 1 u − 4 2 u ⋅ d u = 1 2 ∫ 0 1 ( u u − 4 u ) ⋅ d u = {\displaystyle \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~~~=~~~}
1 2 [ ( u 1 2 + 1 1 2 + 1 ) − ( u − 1 2 + 1 − 1 2 + 1 ) ] = 1 2 [ ( u 3 2 3 2 ) − 4 ( u 1 2 1 2 ) ] = u 3 2 3 − 4 u 1 2 {\displaystyle {\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
![{\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)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/55973ba53e8fe1953589dfb463d33dc794b2d191)