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Question

If dxx23x+2=log(A|+C, then A=.

A
(x32)+x23x+2
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B
(x32)x23x+2
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C
(x+32)+x23x+2
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Solution

The correct option is A (x32)+x23x+2
We can solve this integral as:
I=dxx23x+2I=dxx22.32x+2I=dxx22.32x+9494+2I=dx(x32)2(12)2Substituting t=x32,we get dt=dx, then, our integral becomesI=dtt2(12)2We know that dtt2a2=log(t+t2a2+C,Thus, I=log(t+t214+CNow, substituting back t we get,I=log((x32)+x23x+2+C
Thus, comparing we get A=(x32)+x23x+2

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