How do you find the antiderivative of #x^2*e^(2x)#?

Question

Paisley Johnson · Accepted Answer

#= 1/2 e^(2x) (x^2 - x + 1/2 ) + C#

Employ Integration by Components

#int x^2 e^(2x) dx = int x^2 (1/2 e^(2x))^prime dx #

#= 1/2 x^2 e^(2x) - int (x^2)^prime * 1/2 e^(2x) dx #

#= 1/2 x^2 e^(2x) - int x e^(2x) dx #

#= 1/2 x^2 e^(2x) - int x (1/2 e^(2x))^prime dx #

#= 1/2 x^2 e^(2x) - ( 1/2 x e^(2x) - int (x)^prime * 1/2 e^(2x) dx )#

#= 1/2 x^2 e^(2x) - 1/2 x e^(2x) + int 1/2 e^(2x) dx #

#= 1/2 x^2 e^(2x) - 1/2 x e^(2x) + 1/4 e^(2x) + C#

#= 1/2 e^(2x) (x^2 - x + 1/2 ) + C#

Scarlett Allison · Answer

undefined To find the antiderivative of $ x^2 e^{2x} $, you can use integration by parts.

Let:
$$ u = x^2 $$
$$ dv = e^{2x} dx $$

Then, differentiate $ u $ and integrate $ dv $:
$$ du = 2x dx $$
$$ v = \frac{1}{2} e^{2x} $$

Now, use the integration by parts formula:
$$ \int u dv = uv - \int v du $$

Plug in the values:
$$ \int x^2 e^{2x} dx = \frac{1}{2} x^2 e^{2x} - \int \frac{1}{2} e^{2x} \cdot 2x dx $$

Simplify the integral:
$$ \int x^2 e^{2x} dx = \frac{1}{2} x^2 e^{2x} - \int x e^{2x} dx $$

Now, integrate $ \int x e^{2x} dx $ using integration by parts again:
Let:
$$ u = x $$
$$ dv = e^{2x} dx $$

Then, differentiate $ u $ and integrate $ dv $:
$$ du = dx $$
$$ v = \frac{1}{2} e^{2x} $$

Use the integration by parts formula:
$$ \int u dv = uv - \int v du $$

Plug in the values:
$$ \int x e^{2x} dx = \frac{1}{2} x e^{2x} - \int \frac{1}{2} e^{2x} dx $$

Simplify:
$$ \int x e^{2x} dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} $$

Now, substitute this result back into the original expression:
$$ \int x^2 e^{2x} dx = \frac{1}{2} x^2 e^{2x} - \left( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} \right) $$

$$ \int x^2 e^{2x} dx = \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} + C $$

Therefore, the antiderivative of $ x^2 e^{2x} $ is:
$$ \frac{1}{2} x^2 e^{2x} - \frac{1}{2} x e^{2x} + \frac{1}{4} e^{2x} + C $$
where $ C $ is the constant of integration.