Integral of ln x / x^2: Parts, Not Substitution

Parts, with the power as dv

The two factors point in opposite directions. Logarithms get simpler when differentiated and messier when integrated, so $u = \ln x$. The power $x^{-2}$ integrates without effort, so $dv = x^{-2}dx$, giving $du = \frac{dx}{x}$ and $v = -\frac{1}{x}$.

Both terms carry a factor of $-\frac{1}{x}$, which is why the tidy form combines them over a single denominator. Note that the leftover integral $\int x^{-2}dx$ is $-\frac{1}{x}$, so the second minus sign is earned, not a slip.

One power away from a different problem

Set this beside $\int\frac{\ln x}{x}dx$. There the leftover factor $\frac{1}{x}\,dx$ is exactly $d\left(\ln x\right)$, so $u = \ln x$ is a substitution and the answer is $\frac{\left(\ln x\right)^{2}}{2} + C$.

Change the denominator to $x^{2}$ and that link breaks. Now $\frac{1}{x^{2}}dx$ is not the differential of anything in the numerator, so substitution no longer finishes the job on its own and parts takes over. Substituting $u = \ln x$ does turn the integral into $\int ue^{-u}\,du$, but that one still needs parts. The integrands look almost identical; the deciding step is what differs.

The mistakes students make

The first of these is the one examiners see most, because the two integrands sit next to each other in every textbook.

• Reusing the answer to $\int\frac{\ln x}{x}dx$ and writing $\frac{\left(\ln x\right)^{2}}{2} + C$. Its derivative is $\frac{\ln x}{x}$, which is missing a power of $x$.
• Taking $v = \frac{1}{x}$ instead of $-\frac{1}{x}$. That returns $\frac{\ln x + 1}{x} + C$, the exact negative of the correct antiderivative.
• Integrating numerator and denominator separately, as though $\int\frac{\ln x}{x^{2}}dx$ were $\frac{\int \ln x\,dx}{\int x^{2}dx}$. No such rule exists.