How do you integrate #int x^5e^(-x^3)#?

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Answer 1 · 2017-11-16T00:44:44

#int x^5e^(-x^3)dx = -(e^(-x^3)(x^3+1))/3 +C#

Substitute #t=x^3# so that #dt=3x^2dx# and note that:

#x^5dx = x^3/3*3x^2dx = (tdt)/3#

We then have:

#int x^5e^(-x^3)dx = 1/3 int te^(-t)dt#

Now integrate by parts using #t# as integer factor:

#int te^(-t)dt = -int td(e^(-t)) =-te^(-t) + int e^(-t)dt = -e^(-t)(t+1)+C#

and undoing the substitution:

#int x^5e^(-x^3)dx = -(e^(-x^3)(x^3+1))/3 +C#