Answer
$\displaystyle \frac{1}{15}(3x-2)^{5}+C$
Work Step by Step
Shortcut formula:
$\qquad $
$\displaystyle \int(ax+b)^{n}dx=\frac{1}{a}\frac{(ax+b)^{n+1}}{n+1}+C \quad($if $n\neq-1)$
$\displaystyle \int(3x-2)^{4}dx=\quad \left[\begin{array}{l}
a=3\\
b=-2\\
n=4
\end{array}\right]$
Apply the formula:
$=\displaystyle \frac{1}{3}\cdot\frac{(3x-2}{4+1}+C$
= $\displaystyle \frac{1}{15}(3x-2)^{5}+C$
