∫ ( 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 {\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]\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]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ea090c66a0778f82b8b29827b96a136217d02404)