Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 77

$$\frac{729}{12}=\frac{243}{4}$$

Given $$ \int_{0}^{1}(x+1)\left(x^{2}+2 x\right)^{5} d x $$ Let $$ u= x^2+2x \ \ \ \Rightarrow\ \ \ du = (2x+2)dx$$ At $ x=0\to u= 8$ and $ x= 1 \to u= 3$ \begin{aligned} \int_{0}^{1}(x+1)\left(x^{2}+2 x\right)^{5} d x &=\frac{1}{2} \int_{0}^{3} u^{5} d u \\ &=\left.\frac{1}{2}\left(\frac{1}{6} u^{6}\right)\right|_{0} ^{3} \\ &=\frac{1}{12}\left(3^{6}-0^{6}\right) \\ &=\frac{729}{12} \end{aligned}