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∫ee20101x(1+1−lnxlnxln(xlnx)) dx \large \displaystyle\int_{e}^{e^{2010}} \dfrac{1}{x} \left(1+\frac{1-\ln x}{\ln x\ln(\frac{x}{\ln x})} \right) \, dx ∫ee2010x1(1+lnxln(lnxx)1−lnx)dx
The integral above can be expressed as b−ln(a−lna){b-\ln (a- \ln a)}b−ln(a−lna), where aaa and bbb are integers.
Enter a−b{a-b}a−b as your answer.
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