# How do you integrate f(x)=3^xe^(-x+1) using the product rule?

Nov 11, 2016

$\int {e}^{x \ln 3 - x + 1} \mathrm{dx} = {e}^{x \ln 3 - x + 1} / \left(\ln 3 - 1\right)$
We can rewrite it like this$f \left(x\right) = {e}^{x \ln 3} {e}^{- x + 1}$ so that we can write $\int {e}^{x \ln 3} {e}^{- x + 1} \mathrm{dx}$ and integrating by parts we have
$\int {e}^{x \ln 3} {e}^{- x + 1} \mathrm{dx} =$
$= - {e}^{- x + 1} {e}^{x \ln 3} - \int - {e}^{- x + 1} \ln 3 {e}^{x \ln 3} \mathrm{dx} =$
$- {e}^{- x + 1} {e}^{x \ln 3} + \ln 3 \int {e}^{x \ln 3 - x + 1} \mathrm{dx}$
$\int {e}^{x \ln 3 - x + 1} \mathrm{dx} = {e}^{x \ln 3 - x + 1} / \left(\ln 3 - 1\right)$