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

#int x * 2^x dx = (x*2^x)/ln2 - 2^x / (ln2)^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 = (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 #u = x# and #dv = e^(xln2)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, #du = 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^(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 #e^x# is simply #e^x#. So,

#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 #1/ln2# out of the integral, as it is a constant:

#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 #v#. Don't forget the constant of integration:

#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 #2# instead of base #e#:

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