To find the integral [tex]\(\int (\ln x)^{-1} \, dx\)[/tex], we need to approach the problem through the method of substitution.
1. Set Up the Substitution:
Let [tex]\(u = \ln x\)[/tex]. The differential [tex]\(du\)[/tex] can then be derived as follows:
[tex]\[
\frac{du}{dx} = \frac{d(\ln x)}{dx} = \frac{1}{x}
\][/tex]
Thus,
[tex]\[
du = \frac{1}{x} dx \quad \text{or} \quad dx = x \, du.
\][/tex]
2. Substitute:
Since [tex]\(u = \ln x\)[/tex], we know [tex]\(x = e^u\)[/tex]. Substituting in for [tex]\(dx\)[/tex] in terms of [tex]\(du\)[/tex], we get:
[tex]\[
dx = e^u \, du.
\][/tex]
Similarly, substituting [tex]\((\ln x)^{-1}\)[/tex] in terms of [tex]\(u\)[/tex]:
[tex]\[
(\ln x)^{-1} = u^{-1}.
\][/tex]
Therefore, the integral transforms as follows:
[tex]\[
\int (\ln x)^{-1} \, dx = \int u^{-1} \cdot e^u \, du.
\][/tex]
3. Relate to Logarithmic Integral Function:
The integral [tex]\(\int \frac{1}{u} e^u \, du\)[/tex] can be recognized as a particular special function known as the logarithmic integral, often denoted [tex]\(\text{li}(x)\)[/tex].
Thus, we find:
[tex]\[
\int (\ln x)^{-1} \, dx = \text{li}(x).
\][/tex]
The result of the integration is:
[tex]\[
\boxed{\text{li}(x)}.
\][/tex]
This completes the process of finding the integral [tex]\(\int (\ln x)^{-1} \, dx\)[/tex].
