=e0∫1xln(1+1/x)dx
awesome work priyam :)
answer is √(2e). lekin kya woh aaa raha hain idhar?
:):) using the new latex... [1][1]
ln(I)=\int_{0}^{1}{xln(1+\frac{1}{x})dx}
I is the value to be found
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