Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 2

Find the integral $\int x^3 (2 + x^4)^5 dx$, by making $u = 2 + x^4$

If $u = 2 + x^4$, then $du = 4x^3 dx$, so $\displaystyle x^3 dx = \frac{1}{4} du$. And


$
\begin{equation}
\begin{aligned}

\int x^3 (2 + x^4)^5 dx =& \int u ^5 \frac{du}{4}
\\
\\
\int x^3 (2 + x^4)^5 dx =& \frac{1}{4} \int u^5 du
\\
\\
\int x^3 (2 + x^4)^5 dx =& \frac{1}{4} \cdot \frac{u^{5 + 1}}{5 + 1} + C
\\
\\
\int x^3 (2 + x^4)^5 dx =& \frac{1}{4} \cdot \frac{u^6}{6} + C
\\
\\
\int x^3 (2 + x^4)^5 dx =& \frac{u^6}{24} + C
\\
\\
\int x^3 (2 + x^4)^5 dx =& \frac{(2 + x^4)^6}{24} + C



\end{aligned}
\end{equation}
$