Problem 12 - Applied Calculus

Question

Find the integrals. $$\int \frac{\ln x}{x^{2}} d x$$

Step-by-Step Solution

Answer

The integral is \( -\frac{\ln x}{x} + \frac{1}{x} + C \).

Step 1: Identify Parts for Integration by Parts

To solve the integral \( \int \frac{\ln x}{x^2} \, dx \), we will use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Choose \( u = \ln x \) and \( dv = \frac{1}{x^2} \, dx \).

Step 2: Differentiate and Integrate Parts

Differentiate \( u = \ln x \) to find \( du = \frac{1}{x} \, dx \). Integrate \( dv = \frac{1}{x^2} \, dx \) to find \( v = -\frac{1}{x} \).

Step 3: Apply the Integration by Parts Formula

Substitute into the integration by parts formula: \[ \int \frac{\ln x}{x^2} \, dx = \left(-\frac{\ln x}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx.\]

Step 4: Evaluate the Remaining Integral

Now, evaluate \( \int \frac{1}{x^2} \, dx \), which is a standard integral. It is equal to \( -\frac{1}{x} \). Thus, \( \int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \left(-\frac{1}{x}\right) = -\frac{\ln x}{x} + \frac{1}{x} + C \), where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

When dealing with integrals, it's important to distinguish between definite and indefinite integrals. Definite integrals calculate the area under a curve between specific points, while indefinite integrals find the general form of the antiderivative, typically including a constant of integration. The original exercise provided, \( \int \frac{\ln x}{x^{2}} \, dx \), is an indefinite integral, as no limits are specified.In definite integrals, you evaluate the antiderivative at the upper and lower limits and subtract them, thus:

• Calculate \( F(b) \) where \( b \) is the upper limit.
• Calculate \( F(a) \) where \( a \) is the lower limit.
• Perform \( F(b) - F(a) \) to get the exact area under the curve between \( a \) and \( b \).

Integration techniques like substitution or integration by parts can be applied to both definite and indefinite integrals to simplify or solve them. However, only definite integrals provide specific numerical results. Remember to include the constant of integration \( C \) only with indefinite integrals, as it accounts for the family of functions that could represent the antiderivative.

Logarithmic Functions

Logarithmic functions are critical in calculus, especially when dealing with functions involving variables in exponents or products. The natural logarithm function, \( \ln x \), is equivalent to \( \log_e x \), where \( e \) is approximately 2.71828. It's the inverse of the exponential function \( e^x \), making it highly useful in integration.When integrating functions involving \( \ln x \), it can often be beneficial to apply integration by parts. In our exercise, we chose \( u = \ln x \) because differentiating it results in a much simpler function, \( \frac{1}{x} \). This is a key insight when selecting which part of the function to differentiate with integration by parts.Key properties of logarithmic functions:

• \( \ln(ab) = \ln a + \ln b \)
• \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• \( \ln(a^b) = b \cdot \ln a \)

These are instrumental when simplifying complex integrals involving logarithms.

Calculus Formulas

Calculus provides a toolbox of formulas and rules for finding derivatives and integrals. In solving \( \int \frac{\ln x}{x^2} \, dx \), we used the integration by parts formula, which originates from the product rule of differentiation: \( (uv)' = u'v + uv' \).The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \]where \( u \) and \( dv \) are chosen based on ease of differentiation and integration, respectively.The typical process involves:

• Choosing \( u \) such that its derivative \( du \) simplifies the integral.
• Selecting \( dv \) so that integrating \( v \) is straightforward.
• Applying the formula to break the original integral into simpler ones.

Sometimes, the selection might involve trial and error to simplify the problem effectively. For integrals involving logarithms, selecting \( u = \ln x \) is common due to its direct and manageable derivative. Developing a strong intuition for these selections is a crucial skill in calculus.