Question #45e95

1 Answer
sente
Mar 25, 2016

intx^2/(1-x)dx=-x^2/2-x-ln|x-1|+C

Explanation:

Using the property that int(f+g) = intf + intg along with the known integrals intx^ndx = x^(n+1)/(n+1)+C for n!= -1 and int1/xdx = ln|x|+C, we have:

intx^2/(1-x)dx = int(-x-1-1/(x-1))dx

=-intxdx-int1dx-int1/(x-1)dx

=-x^2/2-x-ln|x-1|+C

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