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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (ln (\frac{2 x + 3}{2 x - 3}))

\frac{d}{d x} (e^{x^{3} + 5})

\int x^{- 4} d x

\frac{x ^{2} + 1}{x ^{3} - 1}

x^{2} y^{^{''}} - x y^{^{'}} + y = 2 x

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