$$\int_{1}^{2}{\mathrm dx\over e^x\ln(x+1)}={27\over 100}$$
Can this be possible? The result of rational, it is not supposed to be.
$e^x=1+x+{x^2\over 2!}+{x^3\over 3!}+\cdots$
2026-03-25 04:40:41.1774413641
Rational result, how? $\int_{1}^{2}{\mathrm dx\over e^x\ln(x+1)}={27\over 100}$
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You can't prove that it is $\frac{27}{100}$ because it isn't. It's about $0.2699988\ldots$