How do you integrate #int x^3e^(x^2-1)dx# using integration by parts?

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Answer 1 · 2015-11-09T15:51:18

See the explanation.

#int x^3e^(x^2-1)dx#

Integrateing #x^3# and differentiating #e^(x^2-1)# would give us MORE #x#'s, so let's try the other way around first.

To integrate #e^(x^2-1)#, we'll need another #x# so that we can substitute.

So,
#int x^3e^(x^2-1)dx = int x^2[xe^(x^2-1)dx]#

Let #u = x^2# and #dv = xe^(x^2-1)dx#.

We get #du = 2x dx# and #v = 1/2e^(x^2-1)#.

Our integral becomes

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

The integral may again be evaluated by substitution.

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