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Question

The value of the integral 3x+5x3x2x+1dx is
(where C is integration constant)

A
12lnx+1x1+2(x1)+C
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B
12lnx+1x14(x1)+C
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C
14lnx+1x12(x+1)+C
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D
14lnx+2x14(x1)+C
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Solution

The correct option is B 12lnx+1x14(x1)+C
3x+5x3x2x+1=3x+5(x1)2(x+1)
Let
3x+5(x1)2(x+1)=A(x1)+B(x1)2+C(x+1)3x+5=A(x1)(x+1)+B(x+1)+C(x1)23x+5=A(x21)+B(x+1)+C(x2+12x)(1)
Equating the coefficients of x2,x and constant term, we obtain
A+C=0B2C=3A+B+C=5
On solving, we obtain
B=4, A=12, C=123x+5(x1)2(x+1)=12(x1)+4(x1)2+12(x+1)3x+5(x1)2(x+1)dx=121x1dx+41(x1)2dx+121(x+1)dx=12ln|x1|+4(1x1)+12ln|x+1|+C=12lnx+1x14(x1)+C

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