Problem 66 - Calculus Early Transcendentals

Question

The following integrals require a preliminary step such as long division or a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{2 z^{3}+z^{2}-6 z+7}{z^{2}+z-6} d z$$

Step-by-Step Solution

Answer

Question: Determine the integral of the function $\frac{2z^3 + z^2 - 6z + 7}{z^2 + z - 6}$ with respect to z. Answer: The integral of the given function is $z^2 + z + \frac{19}{5} \ln|z-2| - \frac{28}{5} \ln|z+3| + C$, where $C$ is the constant of integration.

Step 1: Perform Long Division

First, we need to perform long division of the numerator by the denominator like this: $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{2z} & +1 \\ \cline{2-5} z^{2}+z-6 & 2z^{3} & +z^{2} & -6z & +7 \\ \cline{2-3} \multicolumn{2}{r}{-2z^3} & -2z^2 & +12z \\ \cline{2-4} \multicolumn{2}{r}{0} & 3z^2 & -18z & +7\\ \cline{3-4} \multicolumn{2}{r}{\underline{\phantom{xx}}} & -3z^2 & -3z & +18 \\ \cline{3-5} \multicolumn{2}{r}{\underline{\phantom{xx}}} & 0 & 15z & -11 \\ \end{array} $$ So we get: $$ \frac{2 z^{3}+z^{2}-6 z+7}{z^{2}+z-6} = 2z + 1 + \frac{15z-11}{z^2+z-6} $$ Now, we will integrate the decomposed expression: $$ \int \frac{2z^3 + z^2 - 6z + 7}{z^2 + z - 6} dz = \int (2z + 1) dz + \int \frac{15z - 11}{z^2 + z - 6} dz $$

Step 2: Use Partial Fractions Decomposition

We will decompose the fraction into partial fractions: $$ \frac{15z-11}{z^2+z-6} = \frac{A}{z-2} + \frac{B}{z+3} $$ Clear the denominators: $$ 15z - 11 = A(z + 3) + B(z - 2) $$ We can find A and B by the following: Setting z = 2: $$ 15(2) - 11 = A(2 + 3) \\ A = \frac{19}{5} $$ Setting z = -3: $$ 15(-3) - 11 = B(-3 - 2) \\ B = -\frac{28}{5} $$ Now the integral becomes: $$ \int (2z + 1) dz + \int \left(\frac{\frac{19}{5}}{z-2} - \frac{\frac{28}{5}}{z+3}\right) dz $$

Step 3: Integrate Each Term Separately

Now, integrate each term: $$ \begin{aligned} \int (2z + 1) dz &= z^2 + z + C_{1}\\ \int \frac{19}{5(z-2)}dz &= \frac{19}{5} \ln|z-2| + C_2 \\ \int \frac{-28}{5(z+3)}dz &= -\frac{28}{5} \ln|z+3| + C_{3} \end{aligned} $$

Step 4: Combine the Results

Finally, combine the integration results and the constant terms: $$ \int \frac{2 z^{3}+z^{2}-6 z+7}{z^{2}+z-6} d z = z^2 + z + \frac{19}{5} \ln|z-2| - \frac{28}{5} \ln|z+3| + C $$ Here, C is the constant of integration, which is the sum of C₁, C₂, and C₃.

Key Concepts

Partial Fractions

The technique of partial fractions is a way to simplify complex rational functions into simpler fractions that we can easily integrate. Imagine a big complex fraction. Breaking it down into tiny, easy-to-handle fractions is what partial fractions help us do.
When faced with an integral involving a rational function, the first step is often to decompose the fraction into a sum of simpler fractions. For instance, if we have a complex expression in the form \(\frac{15z-11}{z^2+z-6}\), we aim to rewrite this as a sum of fractions like \(\frac{A}{z-2} + \frac{B}{z+3}\).
To do this, follow these steps:

• First, factorize the denominator if it’s not in factored form. In our exercise, \(z^2+z-6\) can be rewritten as \((z-2)(z+3)\).
• The next step is to express the fraction as a sum of fractions with unknown coefficients \(A\) and \(B\). This results in the equation \(15z - 11 = A(z + 3) + B(z - 2)\).
• To find \(A\) and \(B\), substitute convenient values for \(z\) that will allow us to solve these simultaneously. Here, selecting \(z = 2\) and \(z = -3\) gives values for \(A\) and \(B\).

This technique is handy because integrating simpler fractions is straightforward, such as \(\int \frac{A}{z - 2}~dz\). Once decomposition is done, calculating the integral becomes much simpler.

Long Division

Long division is a necessary preliminary step in our original exercise to simplify the integration process. When the degree (the highest power of \(z\)) of the numerator is equal to or greater than the degree of the denominator, dividing can help.
In our exercise, the numerator \(2z^3 + z^2 - 6z + 7\) is a polynomial of degree 3, and the denominator \(z^2 + z - 6\) is a polynomial of degree 2. Because the numerator's degree is higher, we first perform long division. This reduces the complexity of our integral.
Here's how long division works in this context:

• Start by dividing the highest degree term of the numerator by the highest degree term of the denominator. For example, divide \(2z^3\) by \(z^2\), which results in \(2z\).
• Multiply the entire denominator by \(2z\) and subtract the result from the original numerator. This simplifies the expression significantly.
• Repeat this process with the new polynomial that's left until the degree of the remainder is less than the degree of the original denominator.

The result of long division in our problem gives the terms \(2z + 1\), plus a remainder of \(\frac{15z - 11}{z^2 + z - 6}\). With this breakdown, you've streamlined the integral into parts that are much easier to manage.

Change of Variables

A change of variables is an incredibly useful technique in integration, helping to simplify integrands by substituting variables. For the given exercise, while it's not explicitly required, understanding this technique is essential in integration.
This approach involves substituting a new variable for an expression in the integral, usually done to transform a complicated function into something simpler. Consider an integral where a substitution like \(u = g(z)\) could be helpful.
The basic steps for employing change of variables are:

• Identify a part of the integral that can be replaced with a single variable \(u\). This is often done by recognizing a function and its derivative within the integral.
• Substitute the chosen substitution into the integral. Remember to express \(dz\) in terms of \(du\).
• Once the integral is simplified and integrated with respect to \(u\), substitute back the original variable to get your final answer.

For example, if you had an integral involving \(\sqrt{z+3}\), setting \(u = z + 3\), and \(du = dz\), could simplify things greatly. This technique is powerful in calculus, even if not directly used in our current exercise, as it enriches your problem-solving toolkit and makes challenging integrals more approachable.