Lab 04 - Integration of Rational Functions by Partial Fractions.ipynb

Kernel: SageMath 9.2

Lab 04 - Integration of Rational Functions by Partial Fractions

Overview

In this lab, we will explore writing a rational function as a sum of simpler rational functions which can be easily integrated.

Important SageMath Commands Introduced in this Lab

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Related Course Material

Section 8.5

Example 1

We will use SageMath to verify our partial fraction decomposition and our integration of the rational function $f(x) = \dfrac{2}{(x+1)(x-1)}.$ The general form of the partial fraction decomposition of $f(x)$ is $$\dfrac{2}{(x+1)(x-1)} = \dfrac{A}{x+1} + \dfrac{B}{x-1}.$$ In order to solve for $A$ and $B$ by hand, we would clear the denominators by multiplying both sides by the denominator original denominator. This results in the equation $$2 = A(x-1) + B(x+1) = (A + B)x + (-A + B).$$ Setting like terms equal, we obtain the system of linear equations $$\left \{ \begin{array}{rcl} A + B & = & 0 \\ -A + B & = & 2 \end{array} \right .$$ In the previous lab, we used the $\textbf{solve}$ command to solve a single equation. This command will also allow us to solve a system of equations as well.

A,B = var('A','B')     ## We first make A and B variables
eq1 = A + B == 0       ## The first equation in the system
eq2 = -A + B == 2      ## The second equation in the system
solve([eq1, eq2], [A, B]) 

Therefore, we have that the partial fraction decomposition of our rational function is $$\dfrac{2}{(x+1)(x-1)} = -\dfrac{1}{x+1} + \dfrac{1}{x-1}.$$

We can use the command ParseError: KaTeX parse error: Expected 'EOF', got '_' at position 17: …textbf{.partial_̲fraction()} to check our partial fraction decomposition of $f(x)$.

def f(x):
    return 2/((x+1)*(x-1))
show(f(x).partial_fraction())

Therefore, we have that $$\int \dfrac{2}{(x+1)(x-1)} \ dx = \int -\dfrac{1}{x+1} + \dfrac{1}{x-1} \ dx = -\ln|x+1| + \ln|x-1| + C.$$ We can use SageMath to check that our answer is correct by using the $\textbf{integrate}$ command. Remember that SageMath uses $\textbf{log(x)}$ in place of $\textbf{ln|x|}$ and does not include $+ C$ with antiderivatives.

show(integrate(f(x),x))

Example 2

Again, we will use SageMath to help determine the partial fraction decomposition and the antiderivative of the function $f(x) = \dfrac{6x^2 - 7x}{x^3 - 2x^2 + x - 2}$. The first thing we need to do is factor the denominator of $f(x)$. Let $denom(x) = x^3 - 2x^2 + x - 2$ be the denominator of $f(x)$. Recall that $(x-a)$ is a factor of $denom(x)$ if and only if $denom(a) = 0$. Therefore, in order to factor the cubic $denom(x)$, we look for numbers $a$ such that $denom(a) = 0$. Note that $denom(a) = 0$ if and only if $x=a$ is an $x$-intercept of $denom(x)$. Therefore, we use SageMath to plot $denom(x)$ and determine any $x$-intercepts.

def denom(x):
    return x^3 - 2*x^2 + x - 2
plot(denom(x),xmin = -10, xmax = 10)

Using the domain $[-10, 10]$ we see that $denom(x)$ does have an $x$-intercept. However, we can't see exactly what it is since graph is too zoomed out. We can zoom in by choosing a smaller domain.

plot(denom(x), xmin=-3, xmax=3)

Therefore, we see that $x=2$ is an $x$-intercept of $denom(x)$. It follows that $(x-2)$ is a factor of $denom(x)$. Thus, we know that $denom(x) = (x-2)denom_2(x)$ where $denom_2(x)$ is a polynomial. We can find $denom_2(x)$ exactly by dividing $denom(x)$ by $(x-2).$ In order to have SageMath divide the polynomials, we use the ParseError: KaTeX parse error: Expected 'EOF', got '_' at position 14: \textbf{.full_̲simplify()} command.

denom2(x) = denom(x) / (x-2)
denom2(x).full_simplify()

Therefore, $denom(x)$ factors as $(x-2)(x^2+1)$. Note that $x^2 + 1$ is an irreducible quadratic. Thus, the factorization of the denominator of $f(x)$ is $(x-2)(x^2+1)$. It follows that $f(x) = \dfrac{6x^2 - 7x}{(x-2)(x^2+1)}$. Now that we have factored the denominator of $f(x)$, we can use partial fraction decomposition to write $f(x)$ as the sum of simpler fractions. Recall that the general form of $f(x)$ into partial fractions is $$\dfrac{6x^2 - 7x}{(x-2)(x^2+1)} = \dfrac{A}{x-2} + \dfrac{Bx + C}{x^2 + 1}.$$ Multiplying both sides of the equation by $(x-2)(x^2 + 1)$ gives the equation $$6x^2 - 7x = A(x^2 + 1) + (Bx + C)(x-2) = (A+B)x^2 + (C - 2B)x + (A - 2C).$$ Equating coefficients, we obtain the system $$\left \{ \begin{array}{rcl} A + \ \ B \ \ \ \ \ \ \ \ \ \ & = & 6 \\ - \ 2B \ + \ \ C & = & -7 \\ A \ \ \ \ \ \ \ \ \ \ - 2C & = & 0 \end{array} \right .$$ We again use SageMath to solve this system.

A,B,C = var('A','B','C')
eq1 = ## Insert Equation 1 here
eq2 = ## Insert Equation 2 here
eq3 = ## Insert Equation 3 here
solve([eq1, eq2, eq3], [A, B, C])

You should get that $A = 2, B = 4,$ and $C = 1$. Therefore, we have $$\int \dfrac{6x^2 - 7x}{(x-2)(x^2 + 1)} \ dx = \int \dfrac{2}{x-2} \ dx + \int \dfrac{4x + 1}{x^2 + 1} \ dx.$$ The first integral on the right side is simply $2 \ln|x-2|.$ In order to evaluate the second integral, we first need to split the fraction into the sum of two fractions. Note that $$\int \dfrac{4x + 1}{x^2 + 1} \ dx = \int \dfrac{4x}{x^2 + 1} \ dx + \int \dfrac{1}{x^2 + 1} \ dx = 2 \ln|x^2 + 1| + \tan^{-1}(x) + C,$$ where we used $u$-substitution to evaluate the first integral. Therefore, we have that $$\int \dfrac{6x^2 - 7x}{x^3 - 2x^2 + x - 2} \ dx = 2\ln |x-2| + 2\ln|x^2 + 1| + \tan^{-1}(x) + C.$$ Use SageMath to check that this answer is correct.

Example 3

Suppose that we are trying to integrate $f(x) = \dfrac{6x^3 + 50x^2 + 136x + 128}{x^4 + 8x^3 + 18x^2 - 27}.$ Again, we will start by determining the partial fraciton decomposition of $f(x)$. First, we must factor $denom(x) = x^4 + 8x^3 + 18x^2 - 27.$ Let's plot $denom(x)$ and try to determine $x$-intercepts.

def denom(x):
    return x^4 + 8*x^3 + 18*x^2 - 27
plot(denom(x), xmin=-5, xmax=5, ymin = -5, ymax = 5)

We can see that both $x = -3$ and $x = 1$ are $x$-intercepts. Therefore, $denom(x) = (x+3)(x-1)denom_2(x)$ for some polynomial $denom_2(x)$.

denom2(x) = denom(x) / ((x+3)*(x-1))
denom2(x).full_simplify()

Thus, $denom(x) = (x+3)(x-1)(x^2 + 6x + 9)$. We can factor the quadratic $x^2 + 6x + 9$ and obtain $(x+3)^2$. Therefore, we have that $d(x) = (x+3)^3(x-1).$ It follows that $f(x) = \dfrac{6x^3 + 50x^2 + 136x + 128}{(x+3)^3(x-1)}.$ Recall that the general form of the partial fraction of $f(x)$ is $$\dfrac{6x^3 + 50x^2 + 136x + 128}{(x+3)^3(x-1)} = \dfrac{A}{x+3} + \dfrac{B}{(x+3)^2} + \dfrac{C}{(x+3)^3} + \dfrac{D}{(x-1)}.$$ $\textbf{Note:}$ you may not have covered the general form of a repeated linear term like $(x+3)^3$ in your class. When we have a repeated linear, the general form is a constant over each power of the linear term up to the power to which it is being raised.

Continue solving the problem by setting up a system of four linear equations and solving for $A, B, C,$ and $D$. Then, use the partial fraction decomposition to integrate $f(x)$.

Example 4

For each of the rational functions below, first find the partial fraction decomposition of the function by solving a system of linear equations. Then, integrate the function by hand and use SageMath to check that your integration is correct.

1. $f1(x) = \dfrac{7x+4}{x^2 - x - 6}$

2. $f2(x) = \dfrac{5x^2 + 100x - 49}{x^3+ 2x^2 - 71x - 72}$

3. $f3(x) = \dfrac{3x^2 - x + 21}{x^3 + 3x}$

4. $f4(x) = \dfrac{5x^3 - 2x^2 + 4x - 1}{x^4 - x^2}$