∫ ( x + 1 ) 2 x + x 2 d x {\displaystyle \int (x+1){\sqrt {2x+x^{2}}}dx}
u = 2 x + x 2 d u = 2 + 2 x d x 1 2 d u = x + 1 {\displaystyle {\begin{aligned}u&=2x+x^{2}\\[2ex]du&=2+2xdx\\[2ex]{\frac {1}{2}}du&=x+1\end{aligned}}}
∫ ( x + 1 ) 2 x + x 2 d x = 1 2 ∫ u d u = 1 2 ∫ u 1 2 d u = 1 2 ( 2 u 3 2 3 ) + C = 1 3 ( u 3 2 ) + C = 1 3 ( 2 x + x 2 ) 3 2 + C {\displaystyle {\begin{aligned}\int (x+1){\sqrt {2x+x^{2}}}dx&={\frac {1}{2}}\int {\sqrt {u}}du={\frac {1}{2}}\int u^{\frac {1}{2}}du\\[2ex]&={\frac {1}{2}}({\frac {2u^{\frac {3}{2}}}{3}})+C\\[2ex]&={\frac {1}{3}}(u^{\frac {3}{2}})+C\\[2ex]&={\frac {1}{3}}(2x+x^{2})^{\frac {3}{2}}+C\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=2x+x^{2}\\[2ex]du&=2+2xdx\\[2ex]{\frac {1}{2}}du&=x+1\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/457f3b114a8832535d70a7a359fc2dbe034c038b)
![{\displaystyle {\begin{aligned}\int (x+1){\sqrt {2x+x^{2}}}dx&={\frac {1}{2}}\int {\sqrt {u}}du={\frac {1}{2}}\int u^{\frac {1}{2}}du\\[2ex]&={\frac {1}{2}}({\frac {2u^{\frac {3}{2}}}{3}})+C\\[2ex]&={\frac {1}{3}}(u^{\frac {3}{2}})+C\\[2ex]&={\frac {1}{3}}(2x+x^{2})^{\frac {3}{2}}+C\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/64df293c2132ddffb8fda9e21d7f148a4b52200a)