How do you integrate #int x^2e^(x^3)# by parts?

Question

Aurora Alvarado · Accepted Answer

Using integration by parts is very artificial for this integral. Substitution is much more reasonable.

#intx^2e^(x^3) dx# Let #u = x^3#. This makes #du = 3x^2 dx#. The integral becomes #1/3 int e^(x^3) (3x^2dx) = 1/3 int e^u du# # = 1/3 e^u + C# # = 1/3 e^(x^3) + C# If I am told that I must use parts , I'll let #u = 1# and #dv = x^2e^(x^3) dx# so that #du = 0 dx# and #v = 1/3e^(x^3)#. And #uv=int v du = 1 * 1/3e^(x^3) - int 1/3e^(x^3) * 0 du# # = 1/3 e^(x^3) + C#.

Elijah Abram · Answer

undefined To integrate $ \int x^2e^{x^3} $ by parts, let $ u = x^2 $ and $ dv = e^{x^3}dx $.
Then, $ du = 2x \, dx $ and $ v = \frac{1}{3}e^{x^3} $.
Using the integration by parts formula:
$$ \int u \, dv = uv - \int v \, du $$
Substituting the values:
$$ \int x^2e^{x^3} \, dx = \frac{1}{3}x^2e^{x^3} - \int \frac{2}{3}xe^{x^3} \, dx $$
This integral on the right side can be further integrated by letting $ w = x^3 $.
Then, $ dw = 3x^2 \, dx $.
$$ = \frac{1}{3}x^2e^{x^3} - \frac{2}{3}\int e^w \, dw $$
$$ = \frac{1}{3}x^2e^{x^3} - \frac{2}{3}e^{x^3} + C $$
Therefore, $ \int x^2e^{x^3} \, dx = \frac{1}{3}x^2e^{x^3} - \frac{2}{3}e^{x^3} + C $, where $ C $ is the constant of integration.