Understanding the Integral Problem

We are asked to evaluate the difference between two integrals: $\int (\ln x)^{-1} dx - \int (\ln x)^{-2} dx$. This involves powers of the natural logarithm function, $\ln x$. Problems like this often require techniques like integration by parts or recognizing specific derivative patterns.

Applying Integration by Parts

Let's consider the first integral, $\int (\ln x)^{-1} dx$. We can try to use integration by parts. The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

Let's choose:

• $u = (\ln x)^{-1}$
• $dv = dx$

Now, we need to find $du$ and $v$:

• To find $du$, we differentiate $u$ with respect to $x$: $du = \frac{d}{dx} (\ln x)^{-1} = -1 \cdot (\ln x)^{-2} \cdot \frac{d}{dx}(\ln x) = -1 \cdot (\ln x)^{-2} \cdot \frac{1}{x} dx = -(\ln x)^{-2} \frac{1}{x} dx$
• To find $v$, we integrate $dv$: $v = \int dx = x$

Now, substitute $u, v, du,$ and $dv$ into the integration by parts formula:

$\int (\ln x)^{-1} dx = x \cdot (\ln x)^{-1} - \int x \cdot \left(-(\ln x)^{-2} \frac{1}{x}\right) dx$

Simplify the second term:

$\int (\ln x)^{-1} dx = x (\ln x)^{-1} - \int \left(-(\ln x)^{-2}\right) dx$

$\int (\ln x)^{-1} dx = x (\ln x)^{-1} + \int (\ln x)^{-2} dx$

Evaluating the Original Expression

Now, substitute this result back into the original expression we need to evaluate:

$\int (\ln x)^{-1} dx - \int (\ln x)^{-2} dx$

Substitute the result for $\int (\ln x)^{-1} dx$:

$\left( x (\ln x)^{-1} + \int (\ln x)^{-2} dx \right) - \int (\ln x)^{-2} dx$

Notice that the term $\int (\ln x)^{-2} dx$ appears with both a positive and a negative sign. These terms cancel each other out:

$x (\ln x)^{-1} + \int (\ln x)^{-2} dx - \int (\ln x)^{-2} dx = x (\ln x)^{-1}$

Since this is an indefinite integral, we must add the constant of integration, $c$.

The result is $x (\ln x)^{-1} + c$.

Comparing with Options

Let's compare our result with the given options:

• Option 1: $x (\ln x)^{-1} + c$
• Option 2: $x (\ln x)^{-2} + c$
• Option 3: $x (\ln x) + c$
• Option 4: $x (\ln x)^{2} + c$

Our calculated result, $x (\ln x)^{-1} + c$, matches Option 1.

Step-by-Step Solution Summary

1. Identify the given integral difference: $\int (\ln x)^{-1} dx - \int (\ln x)^{-2} dx$.
2. Apply integration by parts to the first integral, $\int (\ln x)^{-1} dx$, by setting $u = (\ln x)^{-1}$ and $dv = dx$.
3. Calculate $du = -(\ln x)^{-2} \frac{1}{x} dx$ and $v = x$.
4. Substitute these into the integration by parts formula: $\int (\ln x)^{-1} dx = x (\ln x)^{-1} - \int x \left(-(\ln x)^{-2} \frac{1}{x}\right) dx$.
5. Simplify the result: $\int (\ln x)^{-1} dx = x (\ln x)^{-1} + \int (\ln x)^{-2} dx$.
6. Substitute this back into the original expression: $\left( x (\ln x)^{-1} + \int (\ln x)^{-2} dx \right) - \int (\ln x)^{-2} dx$.
7. Cancel out the $\int (\ln x)^{-2} dx$ terms.
8. The result is $x (\ln x)^{-1}$.
9. Add the constant of integration, $c$, for the indefinite integral. The final answer is $x (\ln x)^{-1} + c$.

Revision Table: Key Concepts

Additional Information on Integration Techniques

Integration by parts is a fundamental technique, but many other methods exist for solving integrals, depending on the form of the integrand.

• Substitution Method: Used when the integrand contains a function and its derivative. It simplifies the integral by changing the variable.
• Trigonometric Substitution: Applicable for integrands involving expressions like $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$, using trigonometric identities.
• Partial Fraction Decomposition: Used for integrating rational functions (polynomial divided by a polynomial) by breaking the fraction down into simpler parts.
• Integration using Tables: Referencing pre-calculated integrals for common functions.
• Numerical Integration: Methods like the Trapezoidal Rule or Simpson's Rule are used to approximate definite integrals when an analytical solution is difficult or impossible to find.

Choosing the right integration technique is crucial for solving integral problems efficiently.