Step 1: Understanding the Concept:
When dealing with integrals that primarily feature logarithmic functions, a common and powerful technique is to substitute \( t = \ln x \).
This transformation effectively turns a logarithmic problem into an exponential one because \( x = e^t \).
A highly useful pattern in calculus for integrals involving \( e^x \) is:
\[ \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \]
The objective is to manipulate our given integral until it fits this specific template, allowing for an immediate solution through pattern recognition.
Step 2: Key Formula or Approach:
Substitute \( t = \ln x \).
Find the differential: \( dt = \frac{1}{x} dx \implies dx = x dt = e^t dt \).
Express the integral in terms of \( t \) and look for the \( e^t [f(t) + f'(t)] \) pattern.
Step 3: Detailed Explanation:
Let the given integral be \( I = \int \frac{\ln x - 1}{(\ln x)^2} dx \).
Applying the substitution:
Let \( t = \ln x \). This implies \( x = e^t \).
Differentiating \( x \) with respect to \( t \), we get:
\[ \frac{dx}{dt} = e^t \implies dx = e^t dt \]
Now, substitute these into the original integral:
\[ I = \int \frac{t - 1}{t^2} (e^t dt) \]
Rearranging to isolate the \( e^t \) term:
\[ I = \int e^t \left( \frac{t - 1}{t^2} \right) dt \]
Now, let's distribute the denominator \( t^2 \) to each term in the numerator:
\[ I = \int e^t \left( \frac{t}{t^2} - \frac{1}{t^2} \right) dt \]
\[ I = \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) dt \]
We check to see if this matches our standard pattern.
Let \( f(t) = \frac{1}{t} \).
Using the power rule to differentiate:
\[ f'(t) = \frac{d}{dt}(t^{-1}) = -1 \cdot t^{-2} = -\frac{1}{t^2} \]
Our integral perfectly matches the form \( \int e^t [f(t) + f'(t)] dt \).
Applying the rule:
\[ I = e^t f(t) + C \]
\[ I = e^t \left( \frac{1}{t} \right) + C \]
Finally, we must revert to the original variable \( x \) by substituting \( e^t = x \) and \( t = \ln x \):
\[ I = x \left( \frac{1}{\ln x} \right) + C \]
\[ I = \frac{x}{\ln x} + C \]
Step 4: Final Answer:
The result of the integration is \( \frac{x}{\ln x} + C \).
This is found in Option (A).