Integral of 1/(x ln x): Formula, Steps and Examples

The integral of \dfrac{1}{x\ln x} is \ln\lvert\ln x\rvert + C. It is a textbook example of a substitution that is hiding in plain sight: the \frac{1}{x} is the derivative of the \ln x next to it.

The integrand is defined for x > 0 with x \ne 1, since \ln 1 = 0 would make the denominator zero.

Graph of 1/(x ln x)

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

1. Spot the pair. Rewrite the integrand as \frac{1}{\ln x}\cdot\frac{1}{x}. The second factor is the derivative of \ln x.
2. Substitute. Let u = \ln x, so du = \frac{1}{x}\,dx: \int\frac{1}{\ln x}\cdot\frac{1}{x}\,dx = \int\frac{1}{u}\,du
3. Integrate. \int\frac{du}{u} = \ln\lvert u\rvert + C, from the integral of 1/x.
4. Substitute back. \ln\lvert\ln x\rvert + C.

The absolute value matters on 0 < x < 1, where \ln x is negative. For x > 1 you can drop it and write \ln(\ln x).

This is the u-substitution rule at work. Whenever you see \frac{1}{x} multiplied by some function of \ln x, try u = \ln x.

Check by differentiating

Use the chain rule twice:

Definite and improper integrals

From e to e²: \ln(\ln e^2) - \ln(\ln e) = \ln 2 - \ln 1 = \ln 2.

From e to e^e: \ln(e) - \ln(1) = 1. Each step of the form [e^{a}, e^{ea}] adds the same area, 1.

From 2 to ∞: this improper integral diverges.

The growth is painfully slow. Even at b = 10^{100}, \ln(\ln b) \approx \ln 230.3 \approx 5.44. A calculator would never convince you it diverges, but the antiderivative proves it.

The integral test. Because this integral diverges, the integral test says the series \sum_{n=2}^\infty\frac{1}{n\ln n} diverges too, even though its terms shrink faster than \frac{1}{n}.

Worked examples

Example 1: a log squared. \displaystyle\int_2^\infty\frac{dx}{x(\ln x)^2}. With u = \ln x, the antiderivative is -\frac{1}{\ln x}:

This one converges. An extra power of \ln x is enough.

Example 2: log on top. \displaystyle\int\frac{\ln x}{x}\,dx = \int u\,du = \frac{(\ln x)^2}{2} + C.

Example 3: a power of the log. \displaystyle\int\frac{dx}{x\sqrt{\ln x}} = \int u^{-1/2}\,du = 2\sqrt{\ln x} + C, for x > 1.

Example 4: a constant inside. \displaystyle\int\frac{dx}{x\ln(5x)}. Here \frac{d}{dx}\ln(5x) = \frac{1}{x} as well, so the answer is \ln\lvert\ln(5x)\rvert + C.

Example 5: one more layer. \displaystyle\int\frac{dx}{x\ln x\,\ln(\ln x)} = \ln\lvert\ln(\ln x)\rvert + C, using u = \ln(\ln x).

Common mistakes

1. Splitting into \frac{1}{x}\cdot\frac{1}{\ln x} and integrating each factor. You cannot integrate a product factor by factor. \ln x\cdot\ln\lvert\ln x\rvert is wrong.
2. Using integration by parts. It is possible but leads nowhere useful. Substitution finishes in one line.
3. Writing \ln\lvert x\ln x\rvert. That is the antiderivative of \frac{\ln x + 1}{x\ln x}, not of \frac{1}{x\ln x}.
4. Integrating across x = 1. The integrand blows up there, so \int_{1/2}^{2}\frac{dx}{x\ln x} is not a normal integral.

The integral of ln x and the integral of x ln x cover the other common log integrals. For more on this kind of substitution, see the u-substitution guide and the u-substitution practice problems.