How do you integrate #int x^2e^(2x)# by parts?

Question

Emilia Aguilar · Accepted Answer

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

Take #x^2# as finite part, so in the integration by parts the degree of #x# decreases:

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

We can now solve the resulting integral by parts again:

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

and we can now solve the last integral directly:

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

Putting it all together:

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

Cameron Alexander · Answer

undefined To integrate $ \int x^2 e^{2x} $ by parts, you use the integration technique known as integration by parts, which is defined as:

$$ \int u \, dv = uv - \int v \, du $$

Here, you select $ u $ and $ dv $ such that differentiation of $ u $ leads to simplification, and integration of $ dv $ is easy.

Let's choose $ u = x^2 $ and $ dv = e^{2x} \, dx $:

1. Differentiate $ u $ to get $ du $:
   $$ du = 2x \, dx $$

2. Integrate $ dv $ to get $ v $:
   $$ v = \frac{1}{2} e^{2x} $$

Now, apply the integration by parts formula:

$$ \int x^2 e^{2x} \, dx = x^2 \left( \frac{1}{2} e^{2x} \right) - \int \left( \frac{1}{2} e^{2x} \right) (2x \, dx) $$

$$ = \frac{1}{2} x^2 e^{2x} - \int x e^{2x} \, dx $$

Now, we have another integral to evaluate. We can use integration by parts again.

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

1. Differentiate $ u $ to get $ du $:
   $$ du = dx $$

2. Integrate $ dv $ to get $ v $:
   $$ v = \frac{1}{2} e^{2x} $$

Apply the integration by parts formula again:

$$ \int x e^{2x} \, dx = x \left( \frac{1}{2} e^{2x} \right) - \int \left( \frac{1}{2} e^{2x} \right) \, dx $$

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

Now, substitute back into the original equation:

$$ \int x^2 e^{2x} \, dx = \frac{1}{2} x^2 e^{2x} - \left( \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} \right) + C $$

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

So, $ \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 $, where $ C $ is the constant of integration.