How do you integrate #int (x+3)(x-1)^5# using substitution?

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Answer 1 · 2017-01-15T15:57:46

#int (x+3)(x-1)^5dx =1/7(x-1)^6(x+11/3)+C#

You could expand the power of the binomial and integrate using the power rule.

To avoid the burdensome calculation of a fifth power, substitute:

#t= (x-1)#
#dt = dx#

so that:

#int (x+3)(x-1)^5dx = int (t+4)t^5dt = int t^6dt + 4 intt^5dt = t^7/7+ 2/3t^6 +C#

Now substitute back #x#:

#int (x+3)(x-1)^5dx = 1/7(x-1)^7 +2/3(x-1)^6+C = 1/7(x-1)^6(x-1 + 14/3) + C = 1/7(x-1)^6(x+11/3)+C#