Problem 344 - Calculus Volume 1

Question

In the following exercises, verify by differentiation that \(\int \ln x d x=x(\ln x-1)+C, \quad\) then use appropriate changes of variables to compute the integral. $$\int \frac{\ln x}{x^{2}} d x \quad\left(\text {Hint} : \text { Set } u=\frac{1}{x} .\right)$$

Step-by-Step Solution

Answer

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

Step 1: Differentiate the given integral result

We start by differentiating the function that is the result of the integral: \[ x(\ln x - 1) + C. \] To differentiate, we apply the product rule to \(x(\ln x - 1)\). The derivative of \(C\) is zero since it's a constant.

Step 2: Apply the product rule to differentiate

Let \(f(x) = x\) and \(g(x) = \ln x - 1\). According to the product rule: \[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x). \] Differentiate each function:- \(f'(x) = 1\), - \(g'(x) = \frac{d}{dx}(\ln x - 1) = \frac{1}{x}.\)Using the product rule, we find:\[ (x(\ln x - 1))' = 1 \cdot (\ln x - 1) + x \cdot \frac{1}{x} = \ln x - 1 + 1. \]

Step 3: Simplify the result of the differentiation

Simplify the expression obtained from the differentiation:\[ \ln x - 1 + 1 = \ln x. \]This confirms the given result of the integral \(\int \ln x \, dx = x(\ln x - 1) + C\) is indeed correct as its derivative is \(\ln x\).

Step 4: Use substitution for solving the integral

For the integral \( \int \frac{\ln x}{x^2} \, dx \), use the hint to set \( u = \frac{1}{x} \). Then, \( du = -\frac{1}{x^2} \, dx \), so \( dx = -x^2 du \).

Step 5: Substitute and simplify the integral

Substitute \( x = \frac{1}{u} \) and \( dx = -\frac{1}{u^2} \, du \) into the integral:\[ \int \frac{\ln x}{x^2} \, dx = \int \ln \left(\frac{1}{u}\right) \cdot u^2 \cdot \left(-\frac{1}{u^2}\right) \, du. \] Simplify:\[ = -\int ( -\ln u) \, du = \int \ln u \, du. \]

Step 6: Integrate the simplified expression

We know from initial steps that \( \int \ln u \, du = u(\ln u - 1) + C \). Therefore, substituting back \( u = \frac{1}{x} \), we have:\[ -u(\ln u - 1) = -\left(\frac{1}{x}\right)(\ln \left(\frac{1}{x}\right) - 1) = \frac{1}{x}(1 - \ln \frac{1}{x}). \] Simplify to get:\[ \frac{1}{x}(\ln x - 1). \]

Step 7: Ensure limits in substitution are respected

Consider the integration was indefinite, the final result incorporates adding a constant of integration, therefore the complete expression is:\[ \frac{1}{x}(\ln x - 1) + C. \]

Key Concepts

Substitution Method

The substitution method in calculus is a powerful tool for simplifying integrals, especially when dealing with complex functions. Here, we use substitution to transform the given integral into a more manageable form. For the integral \( \int \frac{\ln x}{x^2} \, dx \), the hint suggests setting \( u = \frac{1}{x} \). This substitution is strategic because it transforms the variable and, consequently, simplifies the integration process.

• Start by identifying a part of the integral that can be substituted. In this case, we see \( \frac{1}{x} \) as a candidate.
• Derive the differential: If \( u = \frac{1}{x} \), then differentiate to find \( du = -\frac{1}{x^2} \, dx \). This helps to express \( dx \) in terms of \( du \).

Use this substitution to rewrite the integral entirely in terms of \( u \), which often simplifies it into a standard form that is easier to integrate.

Product Rule

The product rule is an essential formula in calculus for differentiating functions that are the product of two other functions. The formula states that if \( f(x) \) and \( g(x) \) are both differentiable functions, then the derivative of their product is given by:
\[ (f(x)g(x))' = f'(x)g(x) + f(x)g'(x). \]
In our exercise, we need to differentiate \( x(\ln x - 1) \).

• Here, set \( f(x) = x \) which has a derivative \( f'(x) = 1 \).
• Set \( g(x) = \ln x - 1 \) with its derivative as \( g'(x) = \frac{1}{x} \).

Plug these into the product rule formulation to find the derivative of \( x(\ln x - 1) \) as \( \ln x - 1 + 1 = \ln x \). This confirms the differentiation was performed correctly, showing the consistency with the given integral's result.

Logarithmic Integration

Logarithmic integration often arises when dealing with integrals involving logarithmic functions. Integrating \( \ln x \) is a classic example that involves understanding special methods since natural logarithms do not integrate directly into simple algebraic expressions.

• When faced with \( \int \ln x \, dx \), recognize it's more complex than standard polynomial integration.
• Use integration by parts, a technique derived from the product rule, to solve for \( \int \ln x \, dx \). Here, set one function as \( \ln x \) and differentiate it, while setting another as a simple polynomial like \( x = t \) and integrate it.

For integrals like \( \int \frac{\ln x}{x^2} \, dx \), simplify using substitution until it requires basic integration skills, leading to results like \( u(\ln u - 1) \) which could be translated back to the original variable for a final result.

Differentiation Verification

Differentiation verification is an important step in confirming our solution to an integral problem. This involves taking the derivative of the integral's result and checking it against the original integrand. It acts as a way to ensure the work done is correct from start to finish.

• After integrating, e.g., \( x(\ln x - 1) + C \), differentiate it back to check the result matches the original expression \( \ln x \).
• Review each step of the differentiation for errors or omissions, especially focusing on correct application of rules, such as product rule here.

Successfully receiving \( \ln x \) as the derivative reinforces the correctness of the integration method applied originally in the problem. Always conclude this verification with a check against the integral's limits if they're specified or an addition of constant \( C \) for indefinite integrals.