Different results when integrating $1/(x \ln(x))$ partially/by substitution.

Question

By substitution I get $ln(ln(x))$. Partially something completely different:

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

$$\int \frac{1}{x \ln(x)}dx=1+C+\int \frac{1}{x \ln(x)}dx$$

Provided that this is correct, it doesn't really solve the integral - why is that?

Side info: this is not homework.

Answer 1

This method will work for an integral of the form $\displaystyle\int\frac{(\ln x)^r}{x}dx$ if $r\ne-1$, since then

$u=(\ln x)^r, dv=\frac{1}{x}dx$ gives $\displaystyle\int\frac{(\ln x)^r}{x}dx=(\ln x)^{r+1}-r\int\frac{(\ln x)^r}{x}dx$, so

$\displaystyle(1+r)\int\frac{(\ln x)^r}{x}dx=(\ln x)^{r+1}+C$ and so $\displaystyle\int\frac{(\ln x)^r}{x}dx=\frac{1}{r+1}(\ln x)^{r+1}+C$.

(Of course, this is more easily found by substituting $u=\ln x$.)

However, this gives no useful result when $r=-1$, and it does not lead to a contradiction since the integral is only determined up to a constant.

Answer 2

Hint:

Let $\int \frac{1}{x \ln(x)}dx = F(x)$, what you have found is that $F(x)$ and $F(x)+1+C$ has the same derivative $F'(x)=\frac{1}{x \ln(x)}$. But this is obvious.