Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 14

Find the indefinite integral $\displaystyle \int \frac{1}{(5t + 4)^{2.7}} dt$

If we let $u = 5t + 4$, then $du = 5 dx$, so $\displaystyle dx = \frac{1}{5} du$. And


$
\begin{equation}
\begin{aligned}

\int \frac{1}{(5t + 4)^{2.7}} dt =& \int \frac{1}{u^{2.7}} \frac{1}{5} du
\\
\\
\int \frac{1}{(5t + 4)^{2.7}} dt =& \frac{1}{5} \int u^{-2.7} du
\\
\\
\int \frac{1}{(5t + 4)^{2.7}} dt =& \frac{1}{5} \cdot \frac{u^{-2.7 + 1}}{-2.7 + 1} + C
\\
\\
\int \frac{1}{(5t + 4)^{2.7}} dt =& \frac{1}{5} \cdot \frac{u^{-1.7}}{-1.7} + C
\\
\\
\int \frac{1}{(5t + 4)^{2.7}} dt =& -8.5 u^{-1.7} + C
\\
\\
\int \frac{1}{(5t + 4)^{2.7}} dt =& -8.5 (5t + 4)^{-1.7} + C
\\
\\
\int \frac{1}{(5t + 4)^{2.7}} dt =& \frac{-8.5}{(5t + 4)^{1.7}} + C


\end{aligned}
\end{equation}
$