4. $int\frac{dt}{(1 - 6t)^4}$, $u = 1-6t$

Explanation:

Step1: Differentiate $u$ with respect to $t$

Given $u = 1-6t$, then $\frac{du}{dt}=-6$, so $dt=-\frac{1}{6}du$

Step2: Substitute $u$ and $dt$ into the integral

The integral $\int\frac{dt}{(1 - 6t)^4}$ becomes $\int\frac{-\frac{1}{6}du}{u^4}=-\frac{1}{6}\int u^{- 4}du$

Step3: Integrate $u^{-4}$

Using the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=-4$, we have $\int u^{-4}du=\frac{u^{-4 + 1}}{-4+1}=-\frac{1}{3}u^{-3}+C$

Step4: Substitute $u$ back

$-\frac{1}{6}\int u^{-4}du=-\frac{1}{6}\times(-\frac{1}{3}u^{-3})+C=\frac{1}{18u^{3}}+C$ Substituting $u = 1-6t$ back, we get $\frac{1}{18(1 - 6t)^{3}}+C$

Answer:

$\frac{1}{18(1 - 6t)^{3}}+C$