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

\frac{d}{d x} (x e^{x s i n (x)})

\int \frac{sin ( u )}{u} d u

\int \frac{x - 15}{x ^{2} ( x + 3 )} d x

2 y^{^{''}} - 16 y^{^{'}} + 32 y = 4 e^{4 t}

x \to 2 lim (x^{2} - 4 x - 2)

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