Problem 28 - Calculus Early Transcendentals

Question

Use a change of variables to find the following indefinite integrals. Check your work by differentiating. $$\left.\int \frac{x}{x-2} d x \text { (Hint: Let } u=x-2 .\right)$$

Step-by-Step Solution

Answer

Question: Evaluate the indefinite integral \(\int \frac{x}{x-2}\, dx\). Answer: \(\int \frac{x}{x-2}\, dx = (x - 2) + 2\ln|x - 2| + C\)

Step 1: Identify the substitution

Choose a substitution that will simplify the expression inside the integral. In this case, let: $$u = x - 2$$

Step 2: Find the derivative of the new variable with respect to the old variable

Differentiate \(u\) with respect to \(x\): $$\frac{du}{dx} = \frac{d}{dx}(x-2)$$ $$\frac{du}{dx} = 1$$

Step 3: Rearrange to find dx

Multiply by dx on both sides to solve for \(dx\): $$du = dx$$

Step 4: Substitute into the integral and solve

Replace the expression inside the integral with the expressions in terms of the new variable \(u\) and \(du\): $$\int \frac{x}{x-2} dx = \int \frac{x}{u} du$$ Since \(x = u + 2\), make a substitution: $$\int \frac{u+2}{u} du$$ Now, split the fraction: $$\int \left(1+\frac{2}{u}\right)du$$

Step 5: Integrate

Integrate each term with respect to \(u\): $$\int 1\, du + \int \frac{2}{u}\, du$$ $$= u + 2\ln|u| + C$$

Step 6: Substitute back the original variable

Replace \(u\) with its equivalent expression in terms of \(x\): $$= (x - 2) + 2\ln|x - 2| + C$$

Step 7: Check the result by differentiating

Differentiate the resulting expression with respect to \(x\): $$\frac{d}{dx}[(x - 2) + 2\ln|x - 2| + C]$$ The derivative of the first term is \(1\), the second term's derivative is \(\frac{2}{x-2}\), and the constant \(C\) disappears: $$= 1 + \frac{2}{x-2}$$ Combine it as a single fraction: $$= \frac{x}{x-2}$$ The original integrand is recovered, so our solution is correct. The indefinite integral of \(\frac{x}{x-2}\) with respect to \(x\) is given by: $$\int \frac{x}{x-2}\, dx = (x - 2) + 2\ln|x - 2| + C$$

Key Concepts

Integration by Substitution

The technique of integration by substitution is akin to the reverse process of the chain rule in differentiation. It's particularly useful when dealing with complex integrals that can be simplified by changing the variable of integration. The key is to identify a function within the integral that, when replaced by a single variable, reduces the integral to a simpler form.

Here, the substitution used is based on recognizing that the denominator of \frac{x}{x-2}\ can be expressed as a new variable \(u = x - 2\). This allows us to 'transform' the integral into one in terms of \(u\), which is often easier to solve. After the integration is performed in terms of \(u\), the original variable \(x\) is substituted back to complete the solution. Through this method, the integral becomes more manageable and the relationship between the derivative and the integral becomes evident, linking the processes of differentiation and integration.

U-Substitution

Commonly referred to as u-substitution, this is a specific application of integration by substitution. To perform u-substitution, one identifies a portion of the integrand that can be set equal to \(u\), and then computes the differential \(du\) based on the identified portion. It requires solving for \(dx\) in terms of \(du\) to make the appropriate substitutions in the integral.

The goal is to rewrite the entire integral in terms of \(u\), making it straightforward to integrate. After simplification, a direct integration of the terms yields the anti-derivative in terms of \(u\), which subsequently is re-expressed in terms of the original variable. This problem illustrates u-substitution by setting \(u = x - 2\) and consequently simplifying the integration process.

Differentiation

The process of differentiation involves finding the derivative of a function, which represents the rate at which the function is changing at any given point. From a conceptual standpoint, differentiation gives us the slope of the tangent line to a function's graph at any point.

In the context of checking the work for an indefinite integral, differentiation serves as a verification tool. After finding the indefinite integral, differentiating the result should yield the original function before the integration was performed. If it does, this confirms that the integration was executed correctly. In this exercise, the differentiation of the anti-derivative recovers the original integrand, thus confirming the correctness of the integration process.

Natural Log Integration

The natural log integration refers to the integration of functions that result in the natural logarithm, denoted by \(\ln\). In relation to integration by substitution, functions of the form \(\frac{1}{u}\) result in a natural log upon integration: \(\int \frac{1}{u}\, du = \ln|u| + C\), where \(C\) is the constant of integration.

The integral involving a natural log often surfaces when the integrand is a fraction, where the numerator is the derivative of the denominator. This was evident in our example where the \(\frac{2}{u}\) term within the integrand led to the natural logarithmic element in the final solution, \(2\ln|u|\), which after substituting back into \(x\), became \(2\ln|x - 2|\). Learning to recognize when and how the natural log appears in integration is essential for solving a wide variety of integrals.