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Question

The value of the integral 2x(x2+1)(x2+3)dx is
(where C is an arbitrary constant)

A
12lnx2+1x23+C
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B
12lnx21x23+C
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C
12lnx2+1x2+3+C
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D
12lnx21x2+3+C
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Solution

The correct option is C 12lnx2+1x2+3+C
2x(x2+1)(x2+3)
Let x2=t2xdx=dt
I=2x(x2+1)(x2+3)dx =dt(t+1)(t+3)(1)
Let 1(t+1)(t+3)=A(t+1)+B(t+3)
1=A(t+3)+B(t+1)(2)
Equating the coefficients of t and constant, we obtain
A+B=0 and 3A+B=1
On solving, we obtain
A=12 and B=12
1(t+1)(t+3)=12(t+1)12(t+3)I=[12(t+1)12(t+3)]dt =12ln|(t+1)|12ln|t+3|+C =12lnt+1t+3+C =12lnx2+1x2+3+C

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