AP Calculus AB and BC

Integral of 1/(1+e^x): Answer and Trick

Q: What is the integral of 1/(1+e^x)?

It is x - \ln\left(1+e^{x}\right) + C.

Q: What is the trick?

Write the numerator as \left(1+e^{x}\right) - e^{x}, which splits the fraction into 1 minus a logarithmic piece.

The integral of 1 over 1 plus e to the x is x minus the natural log of 1 plus e to the x, plus C. The trick is to write the numerator as 1 plus e to the x minus e to the x, which splits the fraction into 1 minus a piece whose numerator is the denominator's derivative.

\int \frac{1}{1+e^{x}}\,dx = x - \ln\left(1+e^{x}\right) + C

Add and subtract to split it

\frac{1}{1+e^{x}} = \frac{\left(1+e^{x}\right) - e^{x}}{1+e^{x}} = 1 - \frac{e^{x}}{1+e^{x}}

The second piece now has the denominator's derivative on top, so it integrates to a logarithm and the first is trivial.

\int \frac{dx}{1+e^{x}} = x - \ln\left(1+e^{x}\right) + C

The logistic connection

$\frac{1}{1+e^{-x}}$ is the logistic function, and this integrand is its reflection. No absolute value is needed because $1+e^{x}$ is always positive.

Common mistakes

Splitting into $\int 1\,dx$ over $\int (1+e^{x})dx$ . There is no such rule.
Answering $\ln\left(1+e^{x}\right)$ alone, forgetting the leading $x$ .

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Frequently asked questions

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

It is $x - \ln\left(1+e^{x}\right) + C$ .

What is the trick?

Write the numerator as $\left(1+e^{x}\right) - e^{x}$ , which splits the fraction into $1$ minus a logarithmic piece.