Question

Question #45e95

Answer 1

`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`