Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1–40, use the method of partial fraction decomposition to perform the required integration. $$ \int \frac{5 x}{2 x^{3}+6 x^{2}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( -\frac{15}{2x} + \frac{5}{2} \ln|x+3| + C \).

Step 1: Simplify the Integrand

The given integral is \( \int \frac{5x}{2x^3 + 6x^2} \, dx \). First, factor the denominator: \( 2x^3 + 6x^2 = 2x^2(x + 3) \). This simplifies the integrand to \( \frac{5x}{2x^2(x+3)} \).

Step 2: Set Up Partial Fraction Decomposition

To decompose \( \frac{5x}{2x^2(x+3)} \), assume a decomposition of the form: \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+3} \).

Step 3: Form the Equation

Rewrite the expression: \( \frac{5x}{2x^2(x+3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+3} \). Multiply through by the denominator \( 2x^2(x+3) \) to get: \( 5x = A(x)(x+3) + B(x+3) + C(2x^2) \).

Step 4: Solve for Coefficients

Expand and combine like terms to equate coefficients:1. Substitute \(x = 0\) to find \(B\).2. Use other simple substitutions or compare coefficients for \(A\) and \(C\).After simplification through substitutions or comparisons, we find: \( A = 0, B = \frac{15}{2}, C = \frac{5}{2} \).

Step 5: Integrate Each Fraction

Integration of each fraction term separately:- Integrate \( \frac{B}{x^2} \), i.e., \( \int \frac{15}{2x^2} \, dx = -\frac{15}{2x} + C_1 \).- Integrate \( \frac{C}{x+3} \), i.e., \( \int \frac{5}{2(x+3)} \, dx = \frac{5}{2} \ln|x+3| + C_2 \).

Step 6: Combine and Simplify

Combine the integrals: \[ \int \frac{5x}{2x^3 + 6x^2} \, dx = -\frac{15}{2x} + \frac{5}{2} \ln|x+3| + C \] where \( C = C_1 + C_2 \) is the constant of integration.

Key Concepts

Integration Techniques

Integration techniques are essential when solving calculus problems. Sometimes, straightforward methods like direct integration do not work, especially with complex fractions.

That's where partial fraction decomposition comes in handy. It's a method used to break complex rational expressions into simpler parts, making them easier to integrate.

This technique is particularly useful when dealing with rational expressions, which are fractions where both the numerator and denominator are polynomials.

For the given exercise, the integral \( \int \frac{5x}{2x^{3}+6x^{2}} \, dx \) involves breaking down this rational expression into simpler terms using partial fraction decomposition. This allows us to apply basic integration rules, which are more straightforward to evaluate.

• Step 1 involves factoring the denominator to simplify as much as possible.
• Step 2 sets up the partial fractions with coefficients that are determined by equating and solving systems of equations.
• In the final step, these simpler fractions are integrated separately. This approach helps in tackling more complicated integrals with ease.

By mastering these techniques, you can solve a wide range of challenging calculus problems involving integration.

Rational Expressions

Rational expressions, such as \( \frac{5x}{2x^3 + 6x^2} \), play a significant role in calculus problem-solving. These are essentially fractions where both the numerator and the denominator are polynomials.

Before attempting integration, we first try to simplify rational expressions using techniques such as factoring. For instance, the expression in our exercise simplifies as follows:

• Factoring the denominator \( 2x^3 + 6x^2 \) gives \( 2x^2(x+3) \).
• This lets us rewrite the integrand as \( \frac{5x}{2x^2(x+3)} \).

Once simplified, we decompose it into simpler fractions. This is crucial because integration rules often apply to these simpler components, not the complex original expression.

Mastering rational expressions is also critical since they often appear in both college-level calculus and real-world applications, where precise mathematical solutions are required.

Calculus Problem Solving

Solving calculus problems can be a challenging yet rewarding process. To succeed, multiple techniques and concepts like partial fraction decomposition, rational expressions, and integration methods should be utilized efficiently.

In our exercise, solving the integral involves a number of strategic steps:

• **Initial Simplification**: Begin by simplifying the expression. This often alleviates unnecessary complexity. Here, the factorization of the denominator aids in the simplification process.


• **Use of Partial Fractions**: By breaking down the integrand into components, solving the integration becomes manageable. This involves setting up an equation by comparing coefficients and solving for unknowns.


• **Integration**: Each decomposed part is then integrated individually, often resulting in terms involving natural logarithms or simple algebraic functions.

Solving calculus problems with these methods strengthens your understanding and displays how different calculus concepts connect to solve real mathematical challenges.

Building proficiency in this approach equips you with the tools needed for more advanced calculus courses and a variety of scientific fields.