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Question

The value of the integral x(x1)2(x+2)dx
(where m is an arbitrary constant)

A
29lnx2x+4+13(x1)+m
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B
29lnx1x+213(x1)+m
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C
23lnx+1x+213(x1)+m
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D
23lnx1x+313(x1)+m
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Solution

The correct option is B 29lnx1x+213(x1)+m
Let x(x1)2(x+2)=A(x1)+B(x1)2+C(x+2)
x=A(x1)(x+2)+B(x+2)+C(x1)2
By putting x=1;x=2 and x=0, we obtain
A=29, B=13, C=29x(x1)2(x+2)=29(x1)+13(x1)229(x+2)x(x1)2(x+2)dx =291(x1)dx+131(x1)2dx291(x+2)dx =29ln|x1|+13(1x1)29ln|x+2|+m =29lnx1x+213(x1)+m
Where m is an arbitrary constant.

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