Using integration by parts, integrate xe^(2x) dx

Hi Sal,

As you may already know, integration by parts works like this:

Take I = ∫ f(x) dx. Now, write f(x) dx = u . dv, and so you have:

I
= ∫ f(x) dx.
= ∫ u . dv.
= [uv] - ∫ v du.

In your case, you have that:
I = ∫ x e^{2x} dx.

So, take f(x) = x e^{2x} dx. Then, choose u = x, and dv = e^{2x} dx. I have chosen them this way around because when you differentiate x, you get a constant.

Now, by the formula:
I = [uv] - ∫ v du.
{ The square brackets around uv are there because when you have a definite integral, you must apply the limits to the uv term. They are included for completeness. }

We need to calculate v and du.

For v, we must integrate dv:
v
= ∫ dv.
= ∫ e^{2x} dx.
= (1 / 2) e^{2x} + Constant.
{ The constant term will be ignored here and added after the final integration, I'm including it for completeness. }

For du, we must differentiate u, given u=x:
du
= d[u].
= d[x].
= dx.

Finally, we have:
I
= [uv] - ∫ v du
= [x {(1/2) e^{2x}}] - ∫ {(1/2) e^{2x}} dx.
= [x {(1/2) e^{2x}}] - (1/4) e^{2x} + Constant.
= (1/2) x e^{2x} - (1/4) e^{2x} + Constant.
= e^{2x} ((1/2) x - (1/4)) + Constant.
= (1/4) e^{2x} (2x - 1) + Constant.

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Solution:
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I = ∫ x e^{2x} dx.
= (1/4) e^{2x} (2x - 1) + Constant.