Use integration by parts to find integral of e^4x(2x+1) dx

Hi Sal,

For integration by parts, you use:
I = ∫ f(x) dx = ∫ u dv = [uv] - ∫ v du.
{ Square brackets indicate limits of definite integration. }

I can't understand your brackets, but I'm guessing that you have:
f(x) dx = e^{4x} (2x + 1) dx.

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MY METHOD
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Take u = 2x + 1, and dv = e^{4x} dx.

Then du = 2dx, and v = ∫ e^{4x} dx = (1/4) e^{4x} + Constant.

Now, we ignore the constant of integration, because we will add it at the end.

I
= ∫ f(x) dx
= [uv] - ∫ v du.
= [{2x + 1} {(1/4) e^{4x}}] - ∫ {(1/4) e^{4x}} {2dx}.
= [{2x + 1} {(1/4) e^{4x}}] - (1/2) ∫ {e^{4x}} {dx}.
= [{2x + 1} {(1/4) e^{4x}}] - (1/2) (1/4) e^{4x} + Constant.
= (2x + 1) {(1/4) e^{4x}} - (1/8) e^{4x} + Constant.
= (1/8) e^{4x} ((4x + 2) - 1) + Constant.
= (1/8) e^{4x} (4x + 1) + Constant.

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SOLUTION
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I = ∫ e^{4x} (2x + 1) dx.
= (1/8) e^{4x} (4x + 1) + Constant.

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NOTE
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There is no reason you could not expand the brackets, so that you get:
I = 2 ∫ x e^{4x} dx + ∫ e^{4x} dx.

Where you need to do integration by parts on the first term only, as the second is an elementary integral. The result is the same, of course.