How do you integrate # (x^(7))e^(x^(8)) dx#?

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Answer 1 · 2015-04-17T16:19:55

#int x^7e^(x^8) dx#

Use substitution to write this an an integral of #e^u du#

Let #u = x^8#. This makes #du = 8x^7 dx#,

#int x^7e^(x^8) dx = 1/8 int e^(x^8)(8x^7) dx = 1/8 int e^u du#

#=1/8 e^u +C = 1/8 e^(x^8) +C#

(Check by differentiating #1/8 e^(x^8) +C#)

Answer 2 · 2015-04-17T16:27:34

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