How do you integrate $int (3x-2)^9$ using substitution?

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Answer 1 · 2016-12-27T07:47:12

The answer is $=(3x-9)^10/30+C$

We need

$intx^ndx=x^(n+1)/(n+1)+C$

We do this integral by substitution

Let $u=3x-9$ , $=>$ , $du=3dx$

Therefore,

$int(3x-2)^9dx=1/3intu^9du$

$=1/3u^10/10$

$=(3x-9)^10/30+C$

How do you integrate int (3x-2)^9int (3x-2)^9 using substitution?