AP Calculus AB and BC

Integral of ln(x)/x: Answer, Proof, and Steps

Q: Does the answer need absolute value bars?

No. The integrand already requires x > 0 for \ln x to exist, so on the whole domain of the problem \ln x is defined and \frac{(\ln x)^2}{2} + C needs no absolute values.

Q: How is this different from the integral of 1/(x ln x)?

Both use u = \ln x, but the placement of \ln x changes the outcome. Here the integral becomes \int u\,du = \frac{u^2}{2}; there it becomes \int \frac{du}{u} = \ln|u|, giving \ln|\ln x| + C.

The integral of ln(x)/x is (ln x)^2/2 + C. The substitution u = ln x works because du = dx/x, the other factor in the integrand, so the integral collapses to the integral of u du. Differentiating (ln x)^2/2 returns ln(x)/x, which confirms the answer.

\int \frac{\ln x}{x}\,dx = \frac{(\ln x)^2}{2} + C

Substituting u = ln x

The integrand contains $\ln x$ next to $\frac{1}{x}$ , and $\frac{1}{x}$ is the derivative of $\ln x$ . That pairing of a function with its own derivative is the signal for substitution.

u = \ln x, \qquad du = \frac{1}{x}\,dx

Every piece of the integral has a replacement: $\ln x$ becomes $u$ and $\frac{dx}{x}$ becomes $du$ .

\int \frac{\ln x}{x}\,dx = \int u\,du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C

Convert back to x

The answer must be written in the original variable. Stopping at $\frac{u^2}{2} + C$ leaves an undefined letter in the answer and loses credit on the AP exam.

Checking by differentiating

Differentiate the result with the chain rule, treating $(\ln x)^2$ as an outer square applied to $\ln x$ .

\frac{d}{dx}\left(\frac{(\ln x)^2}{2}\right) = \frac{1}{2}\cdot 2\ln x\cdot\frac{1}{x} = \frac{\ln x}{x}

The $\frac{1}{2}$ cancels the $2$ the power rule brings down, and the chain rule supplies the $\frac{1}{x}$ . The integrand comes back exactly, so the antiderivative is right.

The mistakes students make

Dropping the $\frac{1}{2}$ and writing $(\ln x)^2$ . Differentiating that gives $\frac{2\ln x}{x}$ , twice the integrand.
Confusing $\frac{\ln x}{x}$ with $\frac{1}{x\ln x}$ . The second one puts $\ln x$ in the denominator and integrates to $\ln|\ln x| + C$ .
Reaching for integration by parts. It does reach the same answer, but only after the integral reappears on both sides, so substitution is far shorter.
Writing $\ln x^2$ when $(\ln x)^2$ is meant. Those are different functions: $\ln x^2 = 2\ln x$ .

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

Does the answer need absolute value bars?

No. The integrand already requires $x > 0$ for $\ln x$ to exist, so on the whole domain of the problem $\ln x$ is defined and $\frac{(\ln x)^2}{2} + C$ needs no absolute values.

What is the integral of ln(x)/x from 1 to e?

$\int_1^e \frac{\ln x}{x}\,dx = \left[\frac{(\ln x)^2}{2}\right]_1^e = \frac{1}{2} - 0 = \frac{1}{2}$ , since $\ln e = 1$ and $\ln 1 = 0$ .

How is this different from the integral of 1/(x ln x)?

Both use $u = \ln x$ , but the placement of $\ln x$ changes the outcome. Here the integral becomes $\int u\,du = \frac{u^2}{2}$ ; there it becomes $\int \frac{du}{u} = \ln|u|$ , giving $\ln|\ln x| + C$ .