Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 4

Find the integral $\displaystyle \int \frac{dt}{(1 - 6t)^4}$, by making $u = 1 - 6t$

If $u = 1 - 6t$, then $du = -6dt$, so $\displaystyle dt = \frac{-1}{6} du$. And


$
\begin{equation}
\begin{aligned}

\int \frac{1}{(1 - 6t)^4} dt =& \int \frac{1}{u^4} \cdot \frac{-1}{6} du
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\\
\int \frac{1}{(1 - 6t)^4} dt =& \frac{-1}{6} \int ^{-4} du
\\
\\
\int \frac{1}{(1 - 6t)^4} dt =& \frac{-1}{6} \cdot \frac{u^{-4 + 1}}{-4 + 1} + C
\\
\\
\int \frac{1}{(1 - 6t)^4} dt =& \frac{-1}{6} \cdot \frac{u^{-3}}{-3} + C
\\
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\int \frac{1}{(1 - 6t)^4} dt =& \frac{u^{-3}}{18} + C
\\
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\int \frac{1}{(1 - 6t)^4} dt =& \frac{(1 - 6t)^{-3}}{18} + C
\\
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\int \frac{1}{(1 - 6t)^4} dt =& \frac{1}{18 (1 - 6t)^3} + C


\end{aligned}
\end{equation}
$