Forx>0,letf(x)=x∫1logt1+tdt.Thenf(x)+f(1x)is equal to
A
14logx2
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B
14(logx)2
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C
logx
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D
12(logx)2
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Solution
The correct option is D12(logx)2 f(x)=x∫1logt1+tdtf(1x)=1x∫1logt1+tdtPutt=1z⇒dt=−1z2dzf(1x)=x∫1log1z1+1z(−1z2)dz=x∫1logzz(1+z)dz=x∫1logtt(1+t)dt⇒f(x)+f(1x)=x∫1logttdt=(logx)22