=e0∫1xln(1+1/x)dx
4 Answers
Abhishek Priyam
·2009-03-03 03:47:58
:):) using the new latex... [1][1]
ln(I)=\int_{0}^{1}{xln(1+\frac{1}{x})dx}
I is the value to be found
:):) using the new latex... [1][1]
ln(I)=\int_{0}^{1}{xln(1+\frac{1}{x})dx}
I is the value to be found