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

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

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

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

diverges

Solution steps

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

diverges

= diverges

\int_{0}^{1 - x} x - x^{2} - x y d y

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

\int_{0}^{1} ((2 x - x^{2}) - x^{2}) d x

\int_{7}^{2} 1 d x

\int_{0}^{4} \frac{6}{7 x + 5} d x

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