AP Calculus AB and BC

Integral of e^x/(1+e^x): Substitution

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

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

Q: Why no absolute value?

Because 1+e^{x} > 0 for every real x, so the logarithm's argument is never negative.

The integral of e to the x over 1 plus e to the x is the natural log of 1 plus e to the x, plus C. The numerator is exactly the derivative of the denominator, which is the signature of a logarithmic substitution. No absolute value is needed because 1 plus e to the x is always positive.

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

Numerator is the derivative of the denominator

That pattern always produces a logarithm, with no leftover factor.

u = 1+e^{x}, \quad du = e^{x}dx \implies \int \frac{du}{u} = \ln\left|u\right| + C

No bars needed here

Since e^x is always positive, 1 + e^x is always at least 1, so the argument can never be negative. Writing the bars is not wrong, only redundant.

The logistic connection

The integrand is the logistic function $\frac{1}{1+e^{-x}}$ in disguise, which is why this integral turns up in Unit 7 growth models as well as Unit 6.

Common mistakes

Splitting the fraction into $\int e^{x}dx$ over $\int (1+e^{x})dx$ . There is no quotient rule for integrals.
Adding a spare factor. The substitution is exact, so nothing needs compensating.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

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

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

Why no absolute value?

Because $1+e^{x} > 0$ for every real $x$ , so the logarithm's argument is never negative.