∫ 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
( r 2 + 4 ) 3 2 3 − 4 ( r 2 + 4 ) 1 2 + C {\displaystyle {\frac {\left(r^{2}+4\right)^{\frac {3}{2}}}{3}}-4\left(r^{2}+4\right)^{\frac {1}{2}}+C}
∫ 0 1 r 3 4 + r 2 ⋅ d r = [ ( r 2 + 4 ) 3 2 3 − 4 ( r 2 + 4 ) 1 2 ] {\displaystyle \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]}
[ ( 1 2 + 4 ) 3 2 3 − 4 ( 1 2 + 4 ) 1 2 ] − [ ( 0 2 + 4 ) 3 2 3 − 4 ( 0 2 + 4 ) 1 2 ] {\displaystyle \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]}
5 3 2 3 − 4 ( 5 ) 1 2 − ( ( 4 ) 3 2 3 − 8 ) = 5 3 2 3 − 4 5 − 4 3 2 3 − 8 {\displaystyle {\frac {5^{\frac {3}{2}}}{3}}-4\left(5\right)^{\frac {1}{2}}-\left({\frac {\left(4\right)^{\frac {3}{2}}}{3}}-8\right)~~~=~~~{\frac {5^{\frac {3}{2}}}{3}}-4{\sqrt {5}}-{\frac {4^{\frac {3}{2}}}{3}}-8}
![{\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)
![{\displaystyle \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]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/3dd3db678e6c971b123ee12f8b3a3a827bbc49ba)
![{\displaystyle \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]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/351a1c9a2e518cc3dc54d49f41a7a40f778536b4)
