# How do I find the integral intx*2^xdx ?

Aug 3, 2014

$\int x \cdot {2}^{x} \mathrm{dx} = \frac{x \cdot {2}^{x}}{\ln} 2 - {2}^{x} / {\left(\ln 2\right)}^{2} + C$

Process:

This problem will require multiple $u$-substitutions and an application of integration by parts.

First, it's important to know how to integrate exponentials with bases other than $e$. The key is to convert them to base $e$ using laws of exponents.

Here is what I mean:

First, look at the ${2}^{x}$ inside our integral. We know that a logarithm is essentially the exponent required to raise a base to a certain number, so, if we want to rewrite $2$ as an exponential with base $e$, all we need to do is raise $e$ to the natural log of $2$.

In other words,

${2}^{x} = {\left({e}^{\ln} 2\right)}^{x}$.

Using one of the laws of exponents, this is actually the same statement as:

${2}^{x} = \left({e}^{x \ln 2}\right)$.

Using this information, we can rewrite our integral as:

$\int x \cdot {2}^{x} \mathrm{dx} = \int x \cdot {e}^{x \ln 2} \mathrm{dx}$

Now we will perform an integration by parts. Remember the formula:

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

We will let $u = x$ and $\mathrm{dv} = {e}^{x \ln 2} \mathrm{dx}$. Any time you have an integrand with something multiplied by $x$ to the $1$st power, it's a good hint to pick $x$ for $u$.

Therefore, $\mathrm{du} = \mathrm{dx}$. To find $v$, we will need to do a $u$-substitution. Since we already are using the variable $u$ in the integration by parts formula, we will use the letter $q$ instead.

$v = \int {e}^{x \ln 2} \mathrm{dx}$
Let $q = x \ln 2$

Therefore,

$v = \frac{1}{\ln} 2 \int \ln 2 \cdot {e}^{x \ln 2} \mathrm{dx}$
$v = \frac{1}{\ln} 2 \int {e}^{q} \mathrm{dq}$

The integral of ${e}^{x}$ is simply ${e}^{x}$. So,

$v = \frac{1}{\ln} 2 {e}^{q}$
$v = \frac{1}{\ln} 2 {e}^{x \ln 2}$

Now we can start plugging things into our integration by parts formula:

$\int x \cdot {e}^{x \ln 2} \mathrm{dx} = u v - \int v \mathrm{du}$
$\int x \cdot {e}^{x \ln 2} \mathrm{dx} = \frac{x \cdot {e}^{x \ln 2}}{\ln} 2 - \int \frac{1}{\ln} 2 {e}^{x \ln 2} \mathrm{dx}$

We will bring the $\frac{1}{\ln} 2$ out of the integral, as it is a constant:

$\int x \cdot {e}^{x \ln 2} \mathrm{dx} = \frac{x \cdot {e}^{x \ln 2}}{\ln} 2 - \frac{1}{\ln} 2 \int {e}^{x \ln 2} \mathrm{dx}$

Evaluating this second integral should be a breeze, since we immediately notice that it is equal to $v$. Don't forget the constant of integration:

$\int x \cdot {e}^{x \ln 2} \mathrm{dx} = \frac{x \cdot {e}^{x \ln 2}}{\ln} 2 - \frac{{e}^{x \ln 2}}{\ln 2} ^ 2 + C$

Now, it could be a good idea to rewrite all the exponentials with base $2$ instead of base $e$:

$\int x \cdot {2}^{x} \mathrm{dx} = \frac{x \cdot {2}^{x}}{\ln} 2 - {2}^{x} / {\left(\ln 2\right)}^{2} + C$