Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{x^{6}+4 x^{3}+4}{x^{3}-4 x^{2}} d x\)

Step-by-Step Solution

Answer

\(-\frac{1}{4} \ln|x| + \frac{1}{x} + \frac{5}{4} \ln|x-4| + C\).

Step 1: Factor the Denominator

First, factor the denominator \(x^{3} - 4x^{2}\). Notice that you can factor out \(x^{2}\), giving \(x^2(x - 4)\). This simplifies the expression to \(\int \frac{x^{6}+4 x^{3}+4}{x^{2}(x - 4)} dx\).

Step 2: Set Up the Partial Fraction Decomposition

Express \(\frac{x^{6}+4 x^{3}+4}{x^{2}(x-4)}\) as a sum of fractions with unknown coefficients: \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-4}\). Multiply both sides by the denominator \(x^2(x-4)\) to get \(x^6 + 4x^3 + 4 = A(x)(x-4) + B(x-4) + Cx^2\).

Step 3: Solve for the Coefficients

Expand the right side: \(Ax^2 - 4Ax + Bx - 4B + Cx^2\). Equate the coefficients of powers of \(x\) to obtain a system of equations:- \(A + C = 1\) (coefficient of \(x^6\))- \(-4A + B = 0\) (coefficient of \(x^5\))- \(-4B = 4\) (constant term)Solve for \(A\), \(B\), and \(C\). From \(-4B = 4\), we get \(B = -1\). Substituting in \(-4A + B = 0\), gives \(A = -\frac{1}{4}\). Substituting in \(A + C = 1\), we get \(C = \frac{5}{4}\).

Step 4: Rewrite the Integral

Substitute \(A\), \(B\), and \(C\) back into the partial fractions: \(\int \left(\frac{-\frac{1}{4}}{x} + \frac{-1}{x^2} + \frac{\frac{5}{4}}{x-4}\right) dx\). This expression can be rewritten as \(-\frac{1}{4} \int \frac{1}{x} dx - \int \frac{1}{x^2} dx + \frac{5}{4} \int \frac{1}{x-4} dx\).

Step 5: Integrate Each Term

Integrate the terms separately:- \(-\frac{1}{4} \int \frac{1}{x} dx = -\frac{1}{4} \ln|x|\)- \(- \int \frac{1}{x^2} dx = \frac{1}{x}\)- \(\frac{5}{4} \int \frac{1}{x-4} dx = \frac{5}{4} \ln|x-4|\)Add the results: \( -\frac{1}{4} \ln|x| + \frac{1}{x} + \frac{5}{4} \ln|x-4| + C\).

Step 6: Write the Final Result

Combine the integrated terms to yield the integral solution:\[-\frac{1}{4} \ln|x| + \frac{1}{x} + \frac{5}{4} \ln|x-4| + C\] where \(C\) is the constant of integration.

Key Concepts

Understanding Definite Integrals

A definite integral is a way to calculate the area under a curve over a specific interval. Unlike an indefinite integral, which finds a general form of an antiderivative, a definite integral computes a numerical value. Given an integral with specified limits, we solve it by evaluating the antiderivative at these boundaries.
For example, if you have an integral from 1 to 3, you would compute the antiderivative and then substitute 3 and 1 into it, subtracting the latter result from the former. This gives you the total area between the function and the x-axis over the interval from x=1 to x=3.
This basic idea helps in interpreting calculus problems that involve areas and accumulations, and it's essential for problems that involve specific bounds, unlike indefinite integrals that result in a function plus a constant, C.

Exploring Integration Techniques

Integration techniques are methods used to solve integrals that are not straightforward. One such method is partial fraction decomposition. This technique is invaluable when integrating rational functions, which are ratios of polynomials.

• Partial fraction decomposition involves breaking down a complex fraction into simpler fractions that are easier to integrate.
• Once decomposed, each fraction can be integrated using basic rules and techniques such as the power rule, substitution, or natural logarithm derivatives.

Take, for example, the problem solved here. The rational function was decomposed into simpler terms using constants A, B, and C. Then, each term was integrated individually. This approach simplifies complex calculus problems, paving the way for straightforward integration.

Solving Calculus Problems with Partial Fractions

When it comes to calculus problems, especially those involving integration, partial fraction decomposition can be a real lifesaver. It allows us to solve integrals that would otherwise be complicated or unwieldy.
In calculus problems like the one provided, partial fractions make it possible to transform a complex rational expression into a sum of simpler fractions. To do this, we need to:

• Factor the denominator into simpler components.
• Express the original fraction as a sum of fractions, each with one of the simpler components as the denominator.
• Solve for unknown coefficients by equating terms and solving systems of equations.

Ultimately, this method breaks down a challenging calculus problem into manageable parts, allowing each term to be more easily integrated. This not only fosters a deeper understanding of integration but also provides a toolkit for tackling a wide range of calculus problems.