AP Calculus AB and BC

Integral of 1/(x ln x): Answer and Substitution

Q: What is the integral of 1/(x ln x)?

It is \ln\left|\ln x\right| + C. Substitute u = \ln x, so du = \frac{1}{x}\,dx, and the integral becomes \int \frac{1}{u}\,du = \ln\left|u\right| + C.

Q: Why does the answer need an absolute value?

Because \ln x is negative for 0 1.

Q: Is the integral of 1/(x ln x) on the AP Calculus AB exam?

Yes. It is an ordinary u-substitution, Topic 6.9, which is on both AB and BC. The infinite version used in the integral test is BC material.

Q: Does the integral of 1/(x ln x) from 2 to infinity converge?

No, it diverges. The antiderivative \ln\left(\ln x\right) grows without bound, so the limit is infinite. By the integral test this is why \sum \frac{1}{n\ln n} also diverges.

Q: How is this different from the integral of ln x over x?

Both substitute u = \ln x, but \frac{1}{x\ln x} becomes \int u^{-1}\,du = \ln\left|u\right|, while \frac{\ln x}{x} becomes \int u\,du = \frac{u^{2}}{2}. The answers are \ln\left|\ln x\right| + C and \frac{\left(\ln x\right)^{2}}{2} + C.

The integral of 1/(x ln x) with respect to x is ln|ln x| + C. The substitution u = ln x works because du = (1/x) dx is already present in the integrand, turning the problem into the integral of 1/u. The absolute value is required because ln x is negative for x between 0 and 1.

\int \frac{1}{x\ln x}\,dx = \ln\left|\ln x\right| + C

How to integrate 1/(x ln x) by substitution

This integrand is a substitution problem that announces itself. The derivative of $\ln x$ is $\frac{1}{x}$ , and there is a $\frac{1}{x}$ sitting in the integrand already, so the inner function and its derivative are both present.

Write the integrand so the pairing is visible before you substitute.

\int \frac{1}{x\ln x}\,dx = \int \frac{1}{\ln x}\cdot\frac{1}{x}\,dx

Now let $u = \ln x$ , so $du = \frac{1}{x}\,dx$ . Every piece of the integrand is accounted for, with no leftover $x$ to convert.

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

The problem becomes the single most important basic antiderivative on the course.

\int \frac{1}{u}\,du = \ln\left|u\right| + C

Substituting back gives a logarithm of a logarithm, which looks strange but is exactly right.

\int \frac{1}{x\ln x}\,dx = \ln\left|\ln x\right| + C

Check it by differentiating

Differentiate $\ln\left|\ln x\right|$ with the chain rule. The outer derivative is $\frac{1}{\ln x}$ and the inner derivative is $\frac{1}{x}$ , and their product is $\frac{1}{x\ln x}$ , the integrand.

Why the absolute value is not optional

The domain of the integrand is $x > 0$ with $x \neq 1$ , since $\ln 1 = 0$ makes the denominator zero. On $0 < x < 1$ the value of $\ln x$ is NEGATIVE, so writing $\ln\left(\ln x\right)$ without bars would be undefined on half the domain.

Take $x = 0.5$ . Then $\ln 0.5 \approx -0.693$ , and $\ln\left(-0.693\right)$ does not exist, while $\ln\left|-0.693\right| \approx -0.367$ is fine. The absolute value is what lets one formula cover both sides of $x = 1$ .

This is the same reason $\int \frac{1}{x}\,dx = \ln\left|x\right| + C$ carries bars, and it is the most commonly dropped notation in Unit 6.

One antiderivative, two separate intervals

Strictly, $x = 1$ splits the domain into two pieces and the constant may differ on each. AP answers write a single $+C$ , but you cannot integrate ACROSS $x = 1$ : the integrand blows up there, so a definite integral spanning it is improper and in fact divergent.

Where the integral of 1/(x ln x) shows up on the AP exam

Integrating using substitution is Topic 6.9 in Unit 6, on both AB and BC, and this is a standard example of the pattern where the inner function is a logarithm. Unit 6 carries a weighting of 15 to 20 percent on both exams.

A definite integral over an interval that avoids $x = 1$ is straightforward.

\int_{2}^{e^{2}} \frac{1}{x\ln x}\,dx = \left[\ln\left|\ln x\right|\right]_{2}^{e^{2}} = \ln 2 - \ln\left(\ln 2\right)

On BC this integrand is the engine behind a well known convergence result. The integral test compares the series $\sum \frac{1}{n\ln n}$ to $\int_{2}^{\infty} \frac{1}{x\ln x}\,dx$ , and because $\ln\left|\ln x\right|$ grows without bound, the integral diverges and so does the series.

\int_{2}^{\infty} \frac{1}{x\ln x}\,dx = \lim_{b \to \infty}\left[\ln\left(\ln x\right)\right]_{2}^{b} = \infty

That result is worth remembering as a boundary case. The terms $\frac{1}{n\ln n}$ shrink faster than $\frac{1}{n}$ but still not fast enough to converge, which is why a shrinking term is never on its own evidence of convergence.

Common mistakes with the integral of 1/(x ln x)

Writing $\ln\left|x\ln x\right|$ . The rule $\int \frac{1}{u}\,du = \ln\left|u\right|$ requires the DERIVATIVE of $u$ to be present as a factor. Here $u = x\ln x$ has derivative $\ln x + 1$ , which is nowhere in the integrand.
Dropping the inner absolute value and writing $\ln\left(\ln x\right)$ , which is undefined for $0 < x < 1$ .
Choosing $u = x$ , which produces $du = dx$ and changes nothing.
Integrating across $x = 1$ as though the integrand were continuous there. It is not, so any definite integral whose interval contains $1$ is improper and divergent.
Reading the answer as $\left(\ln x\right)^{2}$ territory. That is the antiderivative of $\frac{\ln x}{x}$ , the reciprocal arrangement, which gives $\frac{\left(\ln x\right)^{2}}{2} + C$ .

Two lookalike integrands, two different answers

$\int \frac{1}{x\ln x}\,dx = \ln\left|\ln x\right| + C$ has the logarithm downstairs, while $\int \frac{\ln x}{x}\,dx = \frac{\left(\ln x\right)^{2}}{2} + C$ has it upstairs. Both use $u = \ln x$ ; only the power of $u$ differs, $u^{-1}$ against $u^{1}$ .

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 1/(x ln x)?

It is $\ln\left|\ln x\right| + C$ . Substitute $u = \ln x$ , so $du = \frac{1}{x}\,dx$ , and the integral becomes $\int \frac{1}{u}\,du = \ln\left|u\right| + C$ .

Why does the answer need an absolute value?

Because $\ln x$ is negative for $0 < x < 1$ , and the logarithm of a negative number is undefined. The bars let the single expression $\ln\left|\ln x\right|$ cover the whole domain instead of only $x > 1$ .

Is the integral of 1/(x ln x) on the AP Calculus AB exam?

Yes. It is an ordinary $u$ -substitution, Topic 6.9, which is on both AB and BC. The infinite version used in the integral test is BC material.

Does the integral of 1/(x ln x) from 2 to infinity converge?

No, it diverges. The antiderivative $\ln\left(\ln x\right)$ grows without bound, so the limit is infinite. By the integral test this is why $\sum \frac{1}{n\ln n}$ also diverges.

How is this different from the integral of ln x over x?

Both substitute $u = \ln x$ , but $\frac{1}{x\ln x}$ becomes $\int u^{-1}\,du = \ln\left|u\right|$ , while $\frac{\ln x}{x}$ becomes $\int u\,du = \frac{u^{2}}{2}$ . The answers are $\ln\left|\ln x\right| + C$ and $\frac{\left(\ln x\right)^{2}}{2} + C$ .