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

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Answer 1 · 2016-12-18T01:50:18

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#.