Problem 10 - Calculus: One and Several Variables

Question

Calculate.$$\int \frac{x}{(x+1)(x+2)(x+3)} d x$$.

Step-by-Step Solution

Answer

The short answer is: \(-2 \ln |x+1| + \frac{3}{2} \ln |x+2| + \frac{1}{2} \ln |x+3| + C\)

Step 1: Partial Fraction Decomposition

Let's start by decomposing the fraction: \(\frac{x}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}\)

Step 2: Clear the Fractions

We will get rid of the denominators by multiplying both sides by (x+1)(x+2)(x+3): \(x = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)\)

Step 3: Solve for A, B, and C

Solve for A, B, and C by substituting suitable values for x. Choose x values that will make the expression for all but one of the variables equal to zero: 1. For x = -1, we have -1 = 0 + 0 - 2C, and thus C = 1/2 2. For x = -2, we have -2 = 0 - A(-1)(1) + 0, and thus A = -2 3. For x = -3, we have -3 = B(0)(-2) + 0 + 0, and thus B = 3/2 Now our fraction is decomposed into partial fractions as follows: \(\frac{x}{(x+1)(x+2)(x+3)} = \frac{-2}{x+1} + \frac{3/2}{x+2} + \frac{1/2}{x+3}\)

Step 4: Integrate Term-wise

Now we need to integrate the decomposed fractions term-wise: \(\int \frac{x}{(x+1)(x+2)(x+3)} d x = \int \frac{-2}{x+1} d x + \int \frac{3/2}{x+2} d x + \int \frac{1/2}{x+3} d x\) Recall that the integral of a function \(\frac{1}{ax+b}\) d x is \(\frac{1}{a}\ln|ax+b| + C\), so we can apply this to our integrals: = -2 \(\int \frac{1}{x+1} d x + \frac{3}{2}\) \(\int \frac{1}{x+2} d x + \frac{1}{2}\) \(\int \frac{1}{x+3} d x\) = -2 \(\ln |x+1| + \frac{3}{2} \ln |x+2| + \frac{1}{2} \ln |x+3| + C\) Thus, the final solution is:\[-2 \ln |x+1| + \frac{3}{2} \ln |x+2| + \frac{1}{2} \ln |x+3| + C\]

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus, particularly for integration, where a rational function is expressed as a sum of simpler fractions. This technique is particularly useful when dealing with complex algebraic fractions that don't easily lend themselves to standard integration methods. In our example, we started with a fraction:

• \(\frac{x}{(x+1)(x+2)(x+3)}\)

The goal of decomposition is to rewrite this fraction as a sum of individual fractions where each denominator is a factor from the original denominator. We represented it as:

• \(\frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}\)

By equating coefficients obtained after clearing the fractions, we identified the constants A, B, and C. This step is crucial because it simplifies the integration by breaking it into parts we can integrate independently. It's a systematic approach that's crucial for tackling multi-term denominators.

Integration Techniques

Integration techniques are methods used to compute the integrals of functions. In this exercise, after employing partial fraction decomposition, we proceeded with the integration of each term separately. This is often more manageable:

• \(\int \frac{-2}{x+1} \, dx\)
• \(\int \frac{3/2}{x+2} \, dx\)
• \(\int \frac{1/2}{x+3} \, dx\)

These integrals employ the natural logarithm rule, which states that the integral of \(\frac{1}{x+a}\) is \(\ln |x+a| + C\). For the given fractions, each is a straightforward logarithmic integral:

• First integral: \(-2 \ln |x+1|\)
• Second integral: \(\frac{3}{2} \ln |x+2|\)
• Third integral: \(\frac{1}{2} \ln |x+3|\)

Integration techniques, like using logarithmic rules, allow us to combine individual solutions, delivering a final integrated expression in a clean, concise form.

Definite and Indefinite Integrals

Definite and indefinite integrals are fundamental concepts in calculus. In this solution, we focused on indefinite integration, which results in a family of functions and includes an arbitrary constant \(C\). Indefinite integrals do not evaluate to a specific number. Instead, they describe a set of functions that differentiate to give the integrand. In our problem, the final result was:

• \[-2 \ln |x+1| + \frac{3}{2} \ln |x+2| + \frac{1}{2} \ln |x+3| + C\]

Conversely, definite integrals evaluate the net area under a curve between two limits. While we did not calculate a definite integral here, it's important to note that both forms of integrals are interconnected. Understanding both helps in applications like finding areas and solving differential equations. The constant \(C\) is a key feature of indefinite integrals and accounts for any vertical shifts in the family of antiderivatives.