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

##### 1 Answer

**Process:**

This problem will require multiple

First, it's important to know how to integrate exponentials with bases other than

Here is what I mean:

First, look at the

In other words,

#2^x = (e^ln2)^x# .

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

#2^x = (e^(xln2))# .

Using this information, we can rewrite our integral as:

#int x * 2^x dx = int x * e^(xln2) dx#

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

#int u dv = uv - int v du#

We will let

Therefore,

#v = int e^(xln2)dx#

Let#q = xln 2#

Therefore,

#v = 1/ln2 int ln 2 * e^(x ln 2) dx#

#v = 1/ln2 int e^(q) dq#

The integral of

#v = 1/ln2 e^(q)#

#v = 1/ln2 e^(x ln 2)#

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

#int x*e^(xln2)dx = uv - int v du#

#int x*e^(xln2)dx = (x*e^(x ln 2))/ln2 - int 1/ln2 e^(x ln 2) dx#

We will bring the

#int x*e^(xln2)dx = (x*e^(x ln 2))/ln2 - 1/ln2 int e^(x ln 2) dx#

Evaluating this second integral should be a breeze, since we immediately notice that it is equal to

#int x*e^(xln2)dx = (x*e^(x ln 2))/ln2 - (e^(x ln 2))/(ln2)^2 + C#

Now, it could be a good idea to rewrite all the exponentials with base

#int x * 2^x dx = (x*2^x)/ln2 - 2^x / (ln2)^2 + C#