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Question

Resolve into partial fraction x3−3x−2(x2+x+1)(x+1)2

A
3x1x2x+1+1(x+1)23(x+1)
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B
3x1x2x+1+2(x+1)23(x+1)
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C
3x1x2+x+1+2(x+1)23(x+1)
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D
3x1x2+x+1+2(x+1)2+3(x+1)
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Solution

The correct option is D 3x1x2+x+1+2(x+1)23(x+1)
Let x33x2(x2+x+1)(x+1)2=A(x+1)+B(x+1)2+Cx+D(x2+x+1)
x33x2=A(x+1)(x2+x+1)+B(x2+x+1)+(Cx+D)(x+1)2x33x2=A(x3+2x2+2x+1)+B(x2+x+1)+C(x3+2x2+x)+D(x2+2x+1)
On comparing coefficients
A+C=1,2A+B+2C+D=0,2A+B+C+2D=3,A+B+D=2A=3,B=2,C=3,D=1
Hence
x33x2(x2+x+1)(x+1)2=3(x+1)+2(x+1)2+3x1(x2+x+1)
Hence, option 'C' is correct.

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