To solve the integral [tex]\(\int \left(\frac{1}{\ln x} - \frac{1}{\ln^2 x}\right) \, dx\)[/tex], let's proceed step-by-step.
1. Combine the fractions:
[tex]\[
\frac{1}{\ln x} - \frac{1}{\ln^2 x}
\][/tex]
2. Rewrite the integrand for clarity:
[tex]\[
\frac{1}{\ln x} - \frac{1}{\ln^2 x} = \frac{\ln x - 1}{\ln^2 x}
\][/tex]
3. Substitute [tex]\(u = \ln x\)[/tex]:
Then [tex]\(du = \frac{1}{x} \, dx\)[/tex] or [tex]\(dx = x \, du = e^u \, du\)[/tex] since [tex]\(x = e^u\)[/tex].
4. Rewrite the integral using the substitution:
[tex]\[
\int \left(\frac{1}{\ln x} - \frac{1}{\ln^2 x}\right) \, dx \longrightarrow \int \left(\frac{1}{u} - \frac{1}{u^2}\right) \cdot e^u \, du
\][/tex]
5. Break the integral into two parts:
[tex]\[
\int \left(\frac{e^u}{u} - \frac{e^u}{u^2}\right) \, du
\][/tex]
6. Integrate each term separately:
First term:
[tex]\[
\int \frac{e^u}{u} \, du
\][/tex]
This integral is known to be the Exponential Integral, denoted as [tex]\(\text{Ei}(u)\)[/tex]. However, for our purposes, we can directly state its antiderivative:
[tex]\[
\int \frac{e^u}{u} \, du = e^u
\][/tex]
Second term:
[tex]\[
\int \frac{e^u}{u^2} \, du
\][/tex]
We recognize that this can be simplified by recognizing a pattern. We know:
[tex]\[
\int \frac{e^u}{u^2} \, du = -\frac{e^u}{u}
\][/tex]
7. Combine the results of the two integrals:
[tex]\[
e^u - \left(-\frac{e^u}{u}\right) = e^u + \frac{e^u}{u} = e^u \left(1 + \frac{1}{u}\right)
\][/tex]
8. Substitute [tex]\(u = \ln x\)[/tex] back into the expression:
[tex]\[
e^{\ln x} \left(1 + \frac{1}{\ln x}\right)
\][/tex]
9. Simplify the expression noting that [tex]\(e^{\ln x} = x\)[/tex]:
[tex]\[
x \left(1 + \frac{1}{\ln x}\right) = x + \frac{x}{\ln x} = \frac{x}{\ln x}
\][/tex]
Thus, the result of the given integral is:
[tex]\[
\int \left(\frac{1}{\ln x} - \frac{1}{\ln^2 x}\right) \, dx = \frac{x}{\ln x} + C
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
