Answer
$-(2x+1) (\dfrac{e^{-3x}}{3}) -\dfrac{2}{9} e^{-3x}+C$
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\dfrac{2x+1}{e^{3x}} \ dx=\int (2x+1) (e^{-3x}) \ dx $
or, $=(2x+1) (-\dfrac{e^{-3x}}{3})-\int (-\dfrac{e^{-3x}}{3})(2) \ dx$
or, $=-(2x+1) (\dfrac{e^{-3x}}{3}) +\dfrac{2}{3} \int e^{-3x} \ dx $
or, $=-(2x+1) (\dfrac{e^{-3x}}{3}) -\dfrac{2}{9} e^{-3x}+C$
