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

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Answer 1 · 2016-07-18T04:39:56

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

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

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

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

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

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