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How do you integrate #int (3-4x)^8# using substitution?

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1 Answer

Jul 18, 2016

#int (3-4x)^(8) dx = -1/36(3-4x)^(9) + C#

Explanation:

In this problem, we can do a u-substitution.

Let #u = 3-4x#, then # du = -4 dx -> -1/4 du = dx#

Rewriting our integral and substituting back our #u# we have

#int (3-4x)^(8) dx = -1/4 int u^8 du =-1/4 int u^8 du #

#= -1/4 * 1/9u^(9) + C#

#= -1/36(3-4x)^(9) + C#

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