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#