What is the integral from-2 to 2 of ((e^x))/(e^x-1) ?

The integral from-2 to 2 of ((e^x))/(e^x-1) is diverges

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integral from-2 to 2 of ((e^x))/(e^x-1)

\int_{- 2}^{2} \frac{( e ^{x} )}{e ^{x} - 1} d x

diverges

Solution steps

\int_{- 2}^{2} \frac{e ^{x}}{e ^{x} - 1} d x

Find undefined (singularity) points:

x = 0

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 2}^{0} \frac{e ^{x}}{e ^{x} - 1} d x + \int_{0}^{2} \frac{e ^{x}}{e ^{x} - 1} d x

\int_{- 2}^{0} \frac{e ^{x}}{e ^{x} - 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{1}^{2} - \frac{1}{2} x + 4 d x

\int_{\frac{π}{4}}^{\frac{3 π}{4}} cos (x) d x

\int_{0}^{x} 64 - t^{2} d t

\int_{2}^{5} ln (5 x) d x

What is the integral from-2 to 2 of ((e^x))/(e^x-1) ?
The integral from-2 to 2 of ((e^x))/(e^x-1) is diverges