# How do you integrate int x*5^x by integration by parts method?

Jun 11, 2017

$\int x \cdot {5}^{x}$ $\mathrm{dx} =$

#### Explanation:

$\int x \cdot {5}^{x}$ $\mathrm{dx}$

Let $f = x$, then $\mathrm{df} = \mathrm{dx}$

And $\mathrm{dg} = {5}^{x}$ $\mathrm{dx}$, then $g = {5}^{x} / \ln 5$

$\int x \cdot {5}^{x}$ $\mathrm{dx} = \frac{x \cdot {5}^{x}}{\ln} 5 - \frac{1}{\ln} 5 \int {5}^{x}$ $\mathrm{dx}$

$= \frac{x \cdot {5}^{x}}{\ln} 5 - {5}^{x} / {\ln}^{2} 5 + \text{c}$

Taking out a factor of $\frac{1}{\ln} ^ 2 5$ yields:

$\int x \cdot {5}^{x}$ $\mathrm{dx} = \frac{1}{\ln} ^ 2 5 \left[\left(x \cdot {5}^{x}\right) \ln 5 - {5}^{x}\right] + \text{c}$

How to integrate ${5}^{x}$ $\mathrm{dx}$

$\int {5}^{x}$ $\mathrm{dx}$

Let $u = {5}^{x}$

Then $\mathrm{du} = {5}^{x} \ln 5$ $\mathrm{dx}$

And $\mathrm{dx} = \frac{1}{{5}^{x} \ln 5}$ $\mathrm{du}$

$\therefore \int {5}^{x}$ $\mathrm{dx} = \int {5}^{x} \times \frac{1}{{5}^{x} \ln 5}$ $\mathrm{du} = \frac{1}{\ln} 5 \int 1$ $\mathrm{du} = \frac{u}{\ln} 5 + \text{c}$
$= {5}^{x} / \ln 5 + \text{c}$