Integrate the following equation by parts.... "integration symbol" x^(3)*e^(2x)?

Answer 1

This will require 3 integration by parts (as x is raised to the third power).
I'll do one to get you started.
Let u = x^3 and dv = e^(2x) dx, then
du = 3x^2 dx and v = (1/2)e^(2x)
uv - integral of (v du)
= (1/2)x^3 e^(2x) - integral of ((3/2)x^2 e^(2x) dx)
Now you try the rest.
Hint: Let u = (3/2)x^2 and dv = e^(2x)
Notice that the power on the x goes down by one with each integration by parts.

Answer 2

Integral (x^3 e^(2x)) dx

Let u = x^3. dv = e^(2x) dx
du = 3x^2 dx v = (1/2)e^(2x)

Therefore, we have

(x^3) (2e^(2x)) - Integral ([3x^2][(1/2)e^(2x)]dx)
2x^3 e^(2x) - Integral ((3/2)x^2 e^(2x))dx

Let's pull the constant (3/2) out of the integral.

2x^3 e^(2x) - (3/2) * Integral (x^2 e^(2x))dx

Now, we solve the new integral by parts.

Let u = x^2. dv = e^(2x) dx
du = 2x dx v = (1/2)e^(2x)

2x^3 e^(2x) - (3/2) [(1/2)x^2 e^(2x) - Integral (x e^(2x))dx]

For simplicity's sake, let's expand the brackets.

2x^3 e^(2x) - (3/4)x^2 e^(2x) + (3/2) Integral (x e^(2x) dx)

Now, we use integration by parts for a third time.

Let u = x. dv = e^(2x) dx
du = dx. v = (1/2)e^(2x)

2x^3 e^(2x) - (3/4)x^2 e^(2x) + (3/2) [(1/2)xe^(2x) - Integral [(1/2)e^(2x) ] dx

Expanding it once more

2x^3 e^(2x) - (3/4)x^2 e^(2x) + (3/4)x e^(2x) - (3/2) Integral [(1/2)e^(2x) ] dx

We can pull the 1/2 out of the integral, to get

2x^3 e^(2x) - (3/4)x^2 e^(2x) + (3/4)x e^(2x) - (3/4) Integral (e^(2x))dx

Now, this is a simple integral to solve; integral of e^(2x) is (1/2)e^(2x), so

2x^3 e^(2x) - (3/4)x^2 e^(2x) + (3/4)x e^(2x) - (3/4)[(1/2)e^(2x)] + C

Simplifying, we get

2x^3 e^(2x) - (3/4)x^2 e^(2x) + (3/4)x e^(2x) - (3/8)e^(2x) + C