Answer
$\displaystyle 4^x [\frac{3x-2}{\ln 4}+\frac{3}{(\ln 4)^2}]+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int(3x-2)4^x \ dx=\displaystyle (3x-2)\frac{4^x}{\ln 4}+ \frac{3}{\ln 4} \int 4^x \ dx$
or, $=\displaystyle (3x-2)\frac{4^x}{\ln 4}+ \frac{3}{\ln 4} (\dfrac{ 4^x}{\ln 4})+C$
or, $=\displaystyle 4^x [\frac{3x-2}{\ln 4}+\frac{3}{(\ln 4)^2}]+C$
