Question

Integral usando o método da substituição

∫ (2x²+2x-3)^10 (2x+1)dx

Answer 1

$I=\int{(2x^{2}+2x-3)^{10}\,(2x+1)\,dx}$


Substituição:

$2x^{2}+2x-3=u\\ \\ \\ \Rightarrow\;\;(4x+2)\,dx=du\\ \\ \Rightarrow\;\;2\cdot (2x+1)\,dx=du\\ \\ \Rightarrow\;\;(2x+1)\,dx=\dfrac{1}{2}\,du$


Substituindo na integral, temos

$I=\int{u^{10}\cdot \dfrac{1}{2}\,du}\\ \\ \\ I=\dfrac{1}{2}\int{u^{10}\,du}\\ \\ \\ I=\dfrac{1}{2}\cdot \dfrac{u^{10+1}}{10+1}+C\\ \\ \\ I=\dfrac{1}{2}\cdot \dfrac{u^{11}}{11}+C\\ \\ \\ I=\dfrac{u^{11}}{22}+C\\ \\ \\ I=\dfrac{1}{22}\cdot (2x^{2}+2x-3)^{11}+C\\ \\ \\ \\ \Rightarrow\;\;\boxed{\begin{array}{c}\int{(2x^{2}+2x-3)^{10}\,(2x+1)\,dx}=\dfrac{1}{22}\cdot (2x^{2}+2x-3)^{11}+C \end{array}}$