To solve the integral [tex]\(\int \frac{1}{x \ln (x)} \, dx\)[/tex], let's proceed step-by-step:
1. Substitution:
We start by making a substitution that can simplify the integrand. Let:
[tex]\[
u = \ln(x)
\][/tex]
Then, the differential [tex]\(du\)[/tex] can be found by differentiating both sides with respect to [tex]\(x\)[/tex]:
[tex]\[
du = \frac{1}{x} \, dx \quad \text{or} \quad dx = x \, du
\][/tex]
Since [tex]\(u = \ln(x)\)[/tex], we can rewrite [tex]\(dx\)[/tex] in terms of [tex]\(u\)[/tex] as:
[tex]\[
dx = x \, du = e^u \, du
\][/tex]
Here, we have used the fact that [tex]\(x = e^u\)[/tex].
2. Rewrite the integral:
Now, substitute [tex]\(u\)[/tex] and [tex]\(dx\)[/tex] into the integral:
[tex]\[
\int \frac{1}{x \ln(x)} \, dx = \int \frac{1}{e^u \cdot u} \cdot e^u \, du
\][/tex]
Notice that the [tex]\(e^u\)[/tex] terms cancel out:
[tex]\[
\int \frac{1}{u} \, du
\][/tex]
3. Integrate:
The integral now is in a simpler form:
[tex]\[
\int \frac{1}{u} \, du = \ln|u| + C
\][/tex]
Since [tex]\(u = \ln(x)\)[/tex], substitute back:
[tex]\[
\ln|\ln(x)| + C
\][/tex]
Given that [tex]\(\ln(x)\)[/tex] is positive for [tex]\(x > 1\)[/tex], we can drop the absolute value for simplicity:
[tex]\[
\ln(\ln(x)) + C
\][/tex]
Thus, the solution to the integral [tex]\(\int \frac{1}{x \ln(x)} \, dx\)[/tex] is:
[tex]\[
\boxed{\ln (\ln (x)) + C}
\][/tex]
