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How do you integrate #int x(x+1)^8# using substitution?

1 Answer

Jan 16, 2017

#(x+1)^10/10-(x+1)^9/9+C#

Explanation:

Substitute #u=x+1# and you get:
#int(u-1)u^8du#
#=int u^9du - int u^8du#
#=u^10/10-u^9/9+C#
Substituting back #u=x+1# gives the answer.

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