# What is the integral of (1+x)3^x?

Mar 11, 2018

$I = \frac{{3}^{x} \left(\ln \left(3\right) x - 1 + \ln \left(3\right)\right)}{\ln \left(3\right)} ^ 2 + C$

#### Explanation:

We want to solve

$I = \int \left(1 + x\right) {3}^{x} \mathrm{dx}$

Rewrite the integrand using ${a}^{x} = {e}^{\ln \left(a\right) x}$

$I = \int {e}^{\ln \left(3\right) x} \mathrm{dx} + \int x {e}^{\ln \left(3\right) x} \mathrm{dx}$

Use integration by parts for the second integral

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let $u = x \implies \mathrm{du} = \mathrm{dx}$

And $\mathrm{dv} = {e}^{\ln \left(3\right) x} \mathrm{dx} \implies v = \frac{1}{\ln} \left(3\right) {e}^{\ln \left(3\right) x}$

$I = \int {e}^{\ln \left(3\right) x} \mathrm{dx} + \frac{1}{\ln} \left(3\right) x {e}^{\ln \left(3\right) x} - \frac{1}{\ln} \left(3\right) \int {e}^{\ln \left(3\right) x} \mathrm{dx}$

$\textcolor{w h i t e}{I} = \frac{1}{\ln} \left(3\right) x {e}^{\ln \left(3\right) x} + \left(1 - \frac{1}{\ln} \left(3\right)\right) \int {e}^{\ln \left(3\right) x} \mathrm{dx}$

$\textcolor{w h i t e}{I} = \frac{1}{\ln} \left(3\right) x {e}^{\ln \left(3\right) x} + \left(1 - \frac{1}{\ln} \left(3\right)\right) \frac{1}{\ln} \left(3\right) {e}^{\ln \left(3\right) x} + C$

$\textcolor{w h i t e}{I} = \frac{{3}^{x} \left(\ln \left(3\right) x - 1 + \ln \left(3\right)\right)}{\ln \left(3\right)} ^ 2 + C$