What is the integral from 0 to infinity of 1/(2+e^x) ?

The integral from 0 to infinity of 1/(2+e^x) is diverges

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integral from 0 to infinity of 1/(2+e^x)

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{2 + e ^{x}} d x = \frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣ + C

Compute the boundaries:

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

diverges

= diverges

\int \frac{1}{8 - 5 x} d x

+ o f + \frac{x - 4}{x + 4}

\int \frac{x ^{5} - 3125}{x - 5} d x

f (x) = 3 x^{2} + 12 x - 36

\int_{- 3}^{1} x^{2} + 2 x - 3 d x

What is the integral from 0 to infinity of 1/(2+e^x) ?
The integral from 0 to infinity of 1/(2+e^x) is diverges