Q.
The value of the integral ∫0∞ xlogx1+x22dx is
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a
7
b
0
c
5log13
d
2log5
answer is B.
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Detailed Solution
∫0∞ xlogx1+x22dx=∫01 xlogx1+x22dx+∫1∞ xlogx1+x22dxPut x=1/y in the second integral, so that dx=−1/y2dy. Ifx→∞ then y→0, and if x=1 then y=1.∴∫1∞ xlogx1+x22dx=∫10 1⋅logy−1y1+1y22−1y2dy=−∫01 ylogy1+y22dy⇒∫0∞ xlogx1+x22dx=∫01 xlogx1+x22dx−∫01 xlogx1+x22dx=0.
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