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}
$