Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 10

Determine the general indefinite integral $\displaystyle \int v\left( v^2 + 2 \right)^2 dv$

$\begin{equation} \begin{aligned} \int v\left( v^2 + 2 \right)^2 dv &= \int v \left( v^4 + 4v^2 + 4 \right) dv\\ \\ \int v\left( v^2 + 2 \right)^2 dv &= \int \left( v^5 + 4v^3 + 4v \right) dv\\ \\ \int v\left( v^2 + 2 \right)^2 dv &= \int v^5 dv + 4 \int v^3 dv + 4 \int vdv\\ \\ \int v\left( v^2 + 2 \right)^2 dv &= \frac{v^{5 + 1}}{5+1} + 4 \left( \frac{v^{3+1}}{3+1} \right) + 4 \left( \frac{v^{1+1}}{1+1} \right) + C\\ \\ \int v\left( v^2 + 2 \right)^2 dv &= \frac{v^6}{6} + \cancel{4} \left( \frac{v^4}{\cancel{4}} \right) + 4 \left( \frac{v^2}{2} \right) + C \\ \\ \int v\left( v^2 + 2 \right)^2 dv &= \frac{v^6}{6} + v^4 + 2v^2 + C \end{aligned} \end{equation}$