∫ 0 a x ⋅ x 2 + a 2 d x {\displaystyle \int _{0}^{a}x\,\cdot {\sqrt {x^{2}+a^{2}}}dx}
u = x 2 + a 2 d u = 2 x d x 1 2 d u = x d x {\displaystyle {\begin{aligned}u&=x^{2}+a^{2}\\[2ex]du&=2xdx\\[2ex]{\frac {1}{2}}du&=xdx\\[2ex]\end{aligned}}}
1 2 ∫ a 2 2 a 2 u ⋅ d u = 1 2 ⋅ a 1 2 + 1 1 2 + 1 | a 2 2 a 2 {\displaystyle {\begin{aligned}{\frac {1}{2}}\int _{a^{2}}^{2a^{2}}{\sqrt {u}}\,\cdot du={\frac {1}{2}}\,\cdot {\frac {a^{{\frac {1}{2}}+1}}{{\frac {1}{2}}+1}}{\bigg |}_{a^{2}}^{2a^{2}}\end{aligned}}}
1 2 ⋅ 2 3 ⋅ u 3 2 | a 2 2 a 2 = 1 3 ( 2 a 2 ) 3 2 − 1 3 ( a 2 ) 3 2 {\displaystyle {\begin{aligned}{\frac {1}{2}}\,\cdot {\frac {2}{3}}\,\cdot \ u^{\frac {3}{2}}{\bigg |}_{a^{2}}^{2a^{2}}={\frac {1}{3}}\left(2a^{2}\right)^{\frac {3}{2}}-{\frac {1}{3}}\left(a^{2}\right)^{\frac {3}{2}}\end{aligned}}}
= 1 3 ⋅ 2 3 2 ⋅ a 3 − 1 3 a 3 = 1 3 ( 2 3 2 − 1 ) a 3 {\displaystyle {\begin{aligned}={\frac {1}{3}}\,\cdot 2^{\frac {3}{2}}\,\cdot {a^{3}}-{\frac {1}{3}}{a^{3}}={\frac {1}{3}}\left(2^{\frac {3}{2}}-1\right){a^{3}}\end{aligned}}}
1 3 ( 2 2 − 1 ) a 3 {\displaystyle {\begin{aligned}{\frac {1}{3}}\left(2{\sqrt {2}}-1\right){a^{3}}\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=x^{2}+a^{2}\\[2ex]du&=2xdx\\[2ex]{\frac {1}{2}}du&=xdx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/b30e1c4f437286276a954ed47d027e3241e5b97c)
