Question #45e95

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Question #45e95

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Answer 1 · 2016-03-25T18:45:34

#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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Question #45e95

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