Problem 61 - Applied Calculus

Question

Find each integral. [Hint: Separate each integral into two integrals, using the fact that the numerator is a sum or difference, and find the two integrals by two different formulas. \(\int \frac{x+1}{x-1} d x\) [Hint: After separating into two integrals find one by a formula and the other by a substitution.

Step-by-Step Solution

Answer

The integral is \(x + \ln|x-1| + C\).

Step 1: Separate the Integral

Given the integral \(\int \frac{x+1}{x-1} \ dx\), first, recognize that the numerator \(x+1\) can be separated into its individual terms: \(x\) and \(1\). This allows us to rewrite the integral as two separate integrals: \(\int \frac{x}{x-1} \ dx + \int \frac{1}{x-1} \ dx\).

Step 2: Simplify the First Integral

Focus on the first integral \(\int \frac{x}{x-1} \ dx\). Notice that the fraction can be simplified by rewriting the integrand: \(\frac{x}{x-1} = 1 + \frac{1}{x-1}\). This transformation simplifies the integral to: \(\int 1 \ dx + \int \frac{1}{x-1} \ dx\).

Step 3: Solve the Integral of a Constant

The integral of the constant 1 is straightforward: \(\int 1 \ dx = x + C_1\), where \(C_1\) is a constant of integration.

Step 4: Solve by Substitution

Now deal with the second integral, \(\int \frac{1}{x-1} \ dx\). Use substitution by letting \(u = x-1\), which gives \(du = dx\). Substitute these into the integral to get \(\int \frac{1}{u} \, du = \ln|u| + C_2\). Replace \(u\) with \(x-1\) to get \(\ln|x-1| + C_2\).

Step 5: Combine the Results

Adding the results from the separate integrals, we obtain the final result: \(x + \ln|x-1| + C\). Merge the constants of integration \(C_1\) and \(C_2\) into a single constant \(C\) for the complete solution.

Key Concepts

Definite Integrals

In Integral Calculus, a definite integral is a fundamental concept that represents the area under a curve between two points on a graph. It tells you how much the function accumulates across an interval. This is different from an indefinite integral, where you find the family of functions whose derivative is the original function you were given.
The notation for a definite integral is \[ \int_a^b f(x) \, dx, \] where \(a\) and \(b\) are the lower and upper limits of integration, respectively. These limits tell you the boundaries within which to calculate the area.

• The fundamental theorem of calculus links the concept of the derivative with the definite integral.
• Evaluating a definite integral often involves finding the antiderivative of the function and then applying the limits to calculate a specific numerical value.

Understanding definite integrals is crucial for problems involving real-world quantities such as displacement, area, and volume in physics and engineering.

Substitution Method

The substitution method is a technique in integration used to simplify the process of finding an integral. This method is particularly useful when you have a composite function that can be transformed into a simpler form.
The basic idea of substitution is to make a substitution of variables that changes the form of the integrand (the function being integrated) to something easier to work with. For example, if you have \( \int f(g(x))g'(x) \, dx, \) the substitution \( u = g(x) \) can lead to the integral being in terms of \( u \), which is often simpler to evaluate.

• Choose the substitution carefully so that the resulting integral is easy to integrate.
• Remember to replace all parts of the differential, including \(dx\), with the new variable's differential, \(du\).
• Always transform back to the original variable once the integration is complete.

Using the substitution method requires practice to determine which part of the integrand should be substituted, but with experience, it becomes an invaluable technique for solving more complex integrals.

Integration Techniques

Integration techniques are various methods used to solve integrals, especially when the functions are not straightforward to integrate directly.
Common techniques include:

• Basic integration formulas: Such as \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), are the foundational tools in integration.
• Substitution method: As previously described, helps with integrals that involve products of functions or more complex compositions.
• Partial fraction decomposition: Used primarily with rational functions, this technique splits one complicated fraction into simpler fractions that are easier to integrate.
• Integration by parts: A method based on the product rule of differentiation, useful for integrating products of functions.

Selecting the appropriate technique often requires some trial and error at first. The more problems you solve, the more intuitive your choice becomes, helping you tackle a wider range of integrals efficiently.