Question

Como devo proceder ?
∫(x+1)√(2x+x²) dx

Answer 1

$I=\int{(x+1)\sqrt{2x+x^{2}\,}dx}$


Substituição:

$u=2x+x^{2}\\ \\ \Rightarrow\;\;du=(2+2x)\,dx\\ \\ \Rightarrow\;\;du=2\,(x+1)\,dx\\ \\ \Rightarrow\;\;(x+1)\,dx=\dfrac{1}{2}\,du$


Substituindo, temos

$I=\int{\sqrt{u}\cdot \dfrac{1}{2}\,du}\\ \\ \\ I=\dfrac{1}{2}\int{\sqrt{u}\,du}\\ \\ \\ I=\dfrac{1}{2}\int{u^{1/2}\,du}\\ \\ \\ I=\dfrac{1}{2}\cdot \dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\\ \\ \\ I=\dfrac{1}{2}\cdot \dfrac{u^{3/2}}{(\frac{3}{2})}+C\\ \\ \\ I=\dfrac{1}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{3}\cdot u^{3/2}+C\\ \\ \\ I=\dfrac{u^{3/2}}{3}+C\\ \\ \\ I=\dfrac{(2x+x^{2})^{3/2}}{3}+C$