Question

Evaluate the indefinite integral using pattern-based substitution: \[ \int \frac{\ln x - 1}{(\ln x)^2}\,dx \]

• A. \( \frac{x}{\ln x} + C \)
• B. \( x\ln x + C \)
• C. \( \frac{\ln x}{x} + C \)
• D. \( \frac{1}{\ln x} + C \)

Hint:

Whenever an integral features \( \ln x \) mixed across fractions, using the substitution \( x = e^t \) will instantly convert a tricky logarithmic problem into a straightforward exponential pattern recognition exercise.

Answer 1

The Correct Option is A

Concept: Integrals containing logarithmic functions can be simplified by substituting \( t = \ln x \). This transforms the expression into a classical exponential template: \[ \int e^t \left[f(t) + f'(t)\right]dt = e^t f(t) + C \]

Step 1: Apply integration by substitution and change the variables. Let \( t = \ln x \). This exponential rewrite gives \( x = e^t \). Differentiating both sides with respect to \( t \): \[ dx = e^t \,dt \] Now, substitute these terms into our original integral: \[ I = \int \frac{t - 1}{t^2} \cdot (e^t \,dt) = \int e^t \left( \frac{t}{t^2} - \frac{1}{t^2} \right) dt \] \[ I = \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) dt \]

Step 2: Match the transformed integral to the standard exponential pattern. Let's define the internal function as \( f(t) = \frac{1}{t} \). Differentiating this function using the power rule: \[ f'(t) = \frac{d}{dt}(t^{-1}) = -1 \cdot t^{-2} = -\frac{1}{t^2} \] Our integral matches the standard pattern perfectly: \( \int e^t \left[f(t) + f'(t)\right]dt \). Evaluating this structural form gives: \[ I = e^t f(t) + C = e^t \left(\frac{1}{t}\right) + C = \frac{e^t}{t} + C \]

Step 3: Substitute original variables back into the solution. Replace \( t \) with \( \ln x \) and \( e^t \) with \( x \) to find the final answer: \[ I = \frac{x}{\ln x} + C \]