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Question

For x>0, let f(x)=x1log t1+t dt. Then f(x)+f(1x) is equal to


A

14(logx)4

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B

log x

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C

12(logx)2

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D

14(logx)2

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Solution

The correct option is C

12(logx)2


f(1x)=1x1log t1+t dt

Put t=1u, so that dt=(1u2) du, then

f(1x)=x1log(1u)1+1u(1u2) du

=x1log uu(1+u) du=x1log tt(1+t) dt

f(x)+f(1x)=x1(1+1t)log t1+t dt

=x1log tt dt

=12 (log t)2]x1=12(log x)2


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