Question

integrate e^x/(e^x-1) from -1 to 1

Answer 1

forget that one. will not be finite

Answer 2

put
\[u=e^x,du=e^xdx\] get
\[\int\frac{du}{u}=\ln(u)=\ln(e^x-1)\]

Answer 3

this is an improper integral, because the path includes 0 and your integrand is not defined at 0. so you have to compute
\[\int_r^1\frac{e^x}{e^x-1}dx\] and then take the limit as r goes to 0

Answer 4

but
\[\lim_{r\rightarrow 0^+} \ln(e^r-1)\] does not exist , so no limit and hence no integral.