How do you integrate #int (x+1)(3x-2)dx#?

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Answer 1 · 2017-02-05T18:07:37

The answer is #=x^3+x^2/2-2x+C#

We need

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

#(x+1)(3x-2)=3x^2+x-2#

Therefore,

#int(x+1)(3x-2)dx=3intx^2dx+intxdx-int2dx#

#=3/3x^3+x^2/2-2x+C#

#=x^3+x^2/2-2x+C#