Single Variable Calculus Questions and Answers

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

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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} $