# Strange error concerning integration by parts

First, this is not homework; I just decided to try a classic integral in a non-standard way and came out with a strange result.

The integral $I:=\int\frac{dx}{x\ln x}$ is well-known to equal $\ln\ln x + C$, and the result is easily obtained with the substitution $u = \ln x$. I tried the alternate method of integration by parts:

$$I=\int\frac{dx}{x\ln x} = \int\frac{1}{\ln x}\frac{dx}{x}$$

Integrating by parts with $u = \frac{1}{\ln x} \implies du = -\frac{dx}{x\ln^2 x}$ and $dv = \frac{dx}{x} \implies v = \ln x$ gives:

$$\int\frac{1}{\ln x}\frac{dx}{x} = \frac{\ln x}{\ln x} - \int \ln x(-\frac{dx}{x\ln^2 x}) = 1 + \int\frac{dx}{x\ln x} = 1 + I$$

This leads to $I=1+I$. Since this is clearly wrong, I'm going to assume I've made a stupid arithmetic mistake somewhere down the line; but after trying it twice I came to this same barrier. What have I done wrong?

• Hint: Constant of integration! Jun 29 '14 at 21:59

## 1 Answer

Nothing; remember that that antiderivatives always have that arbitrary constant $+C$ at the end.

• ...oh. I didn't actually gain any information about the primitive from that, did I? I see. Jun 29 '14 at 22:01
• Nope, none whatsoever. Like so many other things in math, what you've done is absolute Gospel truth, but utterly useless :)
– user61527
Jun 29 '14 at 22:02