AP Calculus AB and BC

Integral of (ln x)^2/x: Substitution

Q: What is the integral of (ln x)^2/x?

It is \frac{\left(\ln x\right)^{3}}{3} + C.

The integral of ln x squared over x is ln x cubed over 3, plus C. The 1 over x is exactly the derivative of ln x, so substituting u equals ln x turns the integrand into u squared and the power rule finishes it.

\int \frac{\left(\ln x\right)^{2}}{x}\,dx = \frac{\left(\ln x\right)^{3}}{3} + C

One substitution, then the power rule

u = \ln x, \quad du = \frac{dx}{x} \implies \int u^{2}du = \frac{u^{3}}{3} + C

The general version is worth carrying: $\int \frac{(\ln x)^{n}}{x}dx = \frac{(\ln x)^{n+1}}{n+1} + C$ for every $n \neq -1$ .

The excluded case

At $n = -1$ the integrand is $\frac{1}{x\ln x}$ and the power rule fails, exactly as it does for $\int x^{-1}dx$ . That case gives $\ln\left|\ln x\right| + C$ instead.

Common mistakes

Reading $\left(\ln x\right)^{2}$ as $\ln\left(x^{2}\right)$ , which is $2\ln x$ and a completely different integral.
Trying integration by parts.
Forgetting the $\frac{1}{3}$ .

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 (ln x)^2/x?

It is $\frac{\left(\ln x\right)^{3}}{3} + C$ .

What is the general rule?

$\int \frac{(\ln x)^{n}}{x}dx = \frac{(\ln x)^{n+1}}{n+1} + C$ for $n \neq -1$ .