How do you integrate #int (5x+4)^5# using substitution?

2 Answers
Douglas K.
Oct 31, 2016

Please see the explanation.

Explanation:

Let #u = 5x + 4#, then #du = 5dx# or dx = #(1/5)du#

#int(5x + 4)^5dx = (1/5)intu^5du = u^6/30 + C#

Reverse the substitution:

#int(5x + 4)^5dx = (5x+4)^6/30 + C#

Douglas K.
Oct 31, 2016

Please see the explanation.

Explanation:

Let #u = 5x + 4#, then #du = 5dx# or dx = #(1/5)du#

#int(5x + 4)^5dx = (1/5)intu^5du = u^6/30 + C#

Reverse the substitution:

#int(5x + 4)^5dx = (5x+4)^6/30 + C#

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