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How do you integrate #(3x + 1)²#?

1 Answer

Oct 24, 2016

#int(3x+1)^2dx=1/9(3x+1)^3+c#

Explanation:

#int(3x+1)^2dx#, Let #u=3x+1#

#du=3dx#, so #dx=(du)/3#

Subbing #u# and #du# into the integral,

#intu^2(du)/3=1/3intu^2du=1/3[u^3/3]+c=1/9u^3+c#

#=1/9(3x+1)^3+c#

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