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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

diverges

= diverges

\int_{- 2}^{5} h (x) d x

\int_{0}^{1} - x^{3} + 4 x^{2} - 3 x d x

\int_{- 5}^{0} (5 + 25 - x^{2}) d x

\int_{5}^{7} 2 x d x

\int_{1}^{8} (x^{2} - 4 x - 12) d x

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