Question

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

Answer 1

`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`