Partial Fraction

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.4 – Integration of Rational Functions by Partial Fractions

Expected Educational Results

• Objective 06–01: I can use partial fractions to evaluate an indefinite integral.

• Objective 06–02: I can use partial fractions to evaluate a definite integral using FTC-II.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Partial Fractions

Method of Partial Fractions

NOTE: The Method of Partial Fractions is only valid for rational functions of the form ${\displaystyle \frac{P(x)}{Q(x)}}$$\displaystyle \frac{P(x)}{Q(x)}$ where

• $P(x)$$P(x)$ and $Q(x)$$Q(x)$ are polynomial functions;

• $\text{degree}(P)<\text{degree}(Q)$$\text{degree}(P)<\text{degree}(Q)$; and

• $Q(x)$$Q(x)$ is factorable into a product of linear and irreducible quadratic factors.

Distinct Linear Factors

Let ${\displaystyle f(x)=\frac{P(x)}{Q(x)}}$$\displaystyle f(x)=\frac{P(x)}{Q(x)}$ be a rational function where $\text{deg}(P)<\text{deg}(Q)$$\text{deg}(P)<\text{deg}(Q)$ and $Q(x)$$Q(x)$ is factorable into distinct linear factors. Then

${\displaystyle \frac{P(x)}{Q(x)}}={\displaystyle \frac{A}{a_{1}x+b_1}}+{\displaystyle \frac{B}{a_{2}x+b_2}}+\cdots$$\dfrac{P(x)}{Q(x)}=\dfrac{A}{a_1x+b_1}+\dfrac{B}{a_2x+b_2}+\cdots$

Quadratic Factors

Let ${\displaystyle f(x)=\frac{P(x)}{Q(x)}}$$\displaystyle f(x)=\frac{P(x)}{Q(x)}$ be a rational function where $\text{deg}(P)<\text{deg}(Q)$$\text{deg}(P)<\text{deg}(Q)$ and $Q(x)$$Q(x)$ is factorable into distinct quadratic factors. Then

${\displaystyle \frac{P(x)}{Q(x)}}={\displaystyle \frac{Ax+B}{a_1{x}^2+b_1}}+{\displaystyle \frac{Cx+D}{a_2{x}^2+b_2}}+\cdots$$\dfrac{P(x)}{Q(x)}=\dfrac{Ax+B}{a_1x^2+b_1}+\dfrac{Cx+D}{a_2x^2+b_2}+\cdots$

Repeated Linear Factor

Let ${\displaystyle f(x)=\frac{P(x)}{Q(x)}}$$\displaystyle f(x)=\frac{P(x)}{Q(x)}$ be a rational function where $\text{deg}(P)<\text{deg}(Q)$$\text{deg}(P)<\text{deg}(Q)$ and $Q(x)$$Q(x)$ is factorable into $r$$r$-repeated linear factors, i.e., $Q(x)=(a_{1}x+b_1{)}^r$$Q(x)=(a_1x+b_1)^r$. Then

${\displaystyle \frac{P(x)}{Q(x)}}={\displaystyle \frac{A}{a_{1}x+b_1}}+{\displaystyle \frac{B}{(a_{1}x+b_1{)}^2}}+{\displaystyle \frac{C}{(a_{1}x+b_1{)}^3}}+\cdots +{\displaystyle \frac{R}{(a_{1}x+b_1{)}^r}}$$\dfrac{P(x)}{Q(x)}=\dfrac{A}{a_1x+b_1}+\dfrac{B}{(a_1x+b_1)^2}+\dfrac{C}{(a_1x+b_1)^3}+\cdots+\dfrac{R}{(a_1x+b_1)^r}$

Repeated Quadratic Factors

Let ${\displaystyle f(x)=\frac{P(x)}{Q(x)}}$$\displaystyle f(x)=\frac{P(x)}{Q(x)}$ be a rational function where $\text{deg}(P)<\text{deg}(Q)$$\text{deg}(P)<\text{deg}(Q)$ and $Q(x)$$Q(x)$ is factorable into $r$$r$-repeated quadratic factors, i.e., $Q(x)=(a_1{x}^2+b_1{)}^r$$Q(x)=(a_1x^2+b_1)^r$. Then

${\displaystyle \frac{P(x)}{Q(x)}}={\displaystyle \frac{Ax+B}{a_1{x}^2+b_1}}+{\displaystyle \frac{Cx+D}{(a_1{x}^2+b_1{)}^2}}+{\displaystyle \frac{Ex+F}{(a_1{x}^2+b_1{)}^3}}+\cdots +{\displaystyle \frac{Sx+T}{(a_1{x}^2+b_1{)}^r}}$$\dfrac{P(x)}{Q(x)}=\dfrac{Ax+B}{a_1x^2+b_1}+\dfrac{Cx+D}{(a_1x^2+b_1)^2}+\dfrac{Ex+F}{(a_1x^2+b_1)^3}+\cdots+\dfrac{Sx+T}{(a_1x^2+b_1)^r}$

NOTE: The above may not be helpful when you first look at the definitions, so, you should try to conceptually understand the Method of Partial Fractions.

Example 01: Express ${\displaystyle \frac{x^2+2x-4}{x^4-1}}$$\displaystyle \frac{x^2+2x-4}{x^4-1}$ as a sum of partial fractions.

Solution 01:

All three conditions needed to use the Method of Partial fractions are met.

Completely, factor the denominator: $x^4-1=(x-1)(x+1)(x^2+1)$$x^4-1=(x-1)(x+1)(x^2+1)$.

There are three (3) factors of the denominator; thus, there must be three (3) partial fractions.

Two of the factors in the denominator are linear factors, so a generic constant, $A$$A$ and $B$$B$, is used for the numerator: $\frac{A}{x-1}$$\frac{A}{x-1}$ and $\frac{B}{x+1}$$\frac{B}{x+1}$.

One of the factors of the denominator is an irreducible quadratic, so a generic linear function is used for the numerator: $\frac{Cx+D}{x^2+1}$$\frac{Cx+D}{x^2+1}$

Thus, the rational expression can be rewritten as the sum of three (3) partial fractions:

${\displaystyle \frac{x^2+2x-4}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}}$$\displaystyle \frac{x^2+2x-4}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}$

Next, we need to solve for the coefficients: $A$$A$, $B$$B$, $C$$C$, and $D$$D$.

There are at least two methods (the traditional textbook method or the method shown below) to solve for the coefficients:

1. Distribute the factors of the denominator to the entire equation and cancel factors:

${\displaystyle (\frac{x^2+2x-4}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1})\left((x-1)(x+1)(x^2+1)\right)}$$\displaystyle \left(\frac{x^2+2x-4}{x^4-1}={\color{brown}\frac{A}{x-1}}+{\color{blue}\frac{B}{x+1}}+{\color{green}\frac{Cx+D}{x^2+1}}\right)\left({\color{red}(x-1)(x+1)(x^2+1)}\right)$

Consider each product:

${\displaystyle \Rightarrow \frac{x^2+2x-4}{\cancel{x^4-1}}\cancel{(x-1)(x+1)(x^2+1)}\Rightarrow x^2+2x-4}$$\displaystyle \Rightarrow\frac{x^2+2x-4}{\cancel{x^4-1}}{\cancel{\color{red}(x-1)(x+1)(x^2+1)}}\Rightarrow x^2+2x-4$, and

${\displaystyle \Rightarrow \frac{A}{\cancel{x-1}}\cancel{(x-1)}(x+1)(x^2+1)\Rightarrow A(x+1)(x^2+1)}$$\displaystyle \Rightarrow{\color{brown}\frac{A}{\cancel{x-1}}}{\color{red}\cancel{(x-1)}(x+1)(x^2+1)}\Rightarrow A(x+1)(x^2+1)$, and

${\displaystyle \Rightarrow \frac{B}{\cancel{x+1}}(x-1)\cancel{(x+1)}(x^2+1)\Rightarrow B(x-1)(x^2+1)}$$\displaystyle \Rightarrow{\color{blue}\frac{B}{\cancel{x+1}}}{\color{red}(x-1)\cancel{(x+1)}(x^2+1)}\Rightarrow B(x-1)(x^2+1)$

${\displaystyle \Rightarrow \frac{Cx+D}{\cancel{x^2+1}}(x-1)(x+1)\cancel{(x^2+1)}\Rightarrow (Cx+d)(x-1)(x+1)}$$\displaystyle \Rightarrow{\color{green}\frac{Cx+D}{\cancel{x^2+1}}}{\color{red}(x-1)(x+1)\cancel{(x^2+1)}}\Rightarrow (Cx+d)(x-1)(x+1)$

Thus,

${\displaystyle \Rightarrow x^2+2x-4=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)}$$\displaystyle\Rightarrow x^2+2x-4=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)$

The traditional method suggests to expand, refactor in terms of powers of $x$$x$, set-up four (4) equations and use algebra to solve for the four (4) unknowns.

2. Alternate method. Do NOT expand.

Let $x=1$${\color{orange}x=1}$. So, ${\displaystyle \Rightarrow 1^2+2(1)-4=A(1+1)(1^2+1)+B{\cancel{(1-1)}}^0(1^2+1)+(C1+D){\cancel{(1-1)}}^0(1+1)}$$\displaystyle\Rightarrow {\color{orange}1}^2+2({\color{orange}1})-4=A({\color{orange}1}+1)({\color{orange}1}^2+1)+B\cancelto{0}{({\color{orange}1}-1)}({\color{orange}1}^2+1)+(C{\color{orange}1}+D)\cancelto{0}{({\color{orange}1}-1)}({\color{orange}1}+1)$

${\displaystyle \Rightarrow -1=A(2)(2)\Rightarrow A=-\frac{1}{4}}$$\displaystyle\Rightarrow -1=A(2)(2)\Rightarrow {\color{orange}A=-\frac{1}{4}}$

Let $x=-1$${\color{green}x=-1}$. So, ${\displaystyle \Rightarrow {(-1)}^2+2(-1)-4=A{\cancel{(-1+1)}}^0({(-1)}^2+1)+B(-1-1)({(-1)}^2+1)+(C(-1)+D)(-1-1){\cancel{(-1+1)}}^0}$$\displaystyle\Rightarrow {\color{green}(-1)}^2+2({\color{green}-1})-4=A\cancelto{0}{({\color{green}-1}+1)}({\color{green}(-1)}^2+1)+B({\color{green}-1}-1)({\color{green}(-1)}^2+1)+(C{\color{green}(-1)}+D)({\color{green}-1}-1)\cancelto{0}{({\color{green}-1}+1)}$

${\displaystyle \Rightarrow -5=B(-2)(2)\Rightarrow B=\frac{5}{4}}$$\displaystyle\Rightarrow -5=B(-2)(2)\Rightarrow {\color{green}B=\frac{5}{4}}$

Let $x=0$${\color{pink}x=0}$. So, ${\displaystyle \Rightarrow 0^2+2(0)-4=-\frac{1}{4}(0+1)(0^2+1)+\frac{5}{4}(0-1)(0^2+1)+(C(0)+D)(0-1)(0+1)}$$\displaystyle\Rightarrow {\color{pink}0}^2+2({\color{pink}0})-4={\color{orange}-\frac{1}{4}}({\color{pink}0}+1)({\color{pink}0}^2+1)+{\color{green}\frac{5}{4}}({\color{pink}0}-1)({\color{pink}0}^2+1)+(C{\color{pink}(0)}+D)({\color{pink}0}-1)({\color{pink}0}+1)$

${\displaystyle \Rightarrow -4=-\frac{1}{4}-\frac{5}{4}-D\Rightarrow D=\frac{5}{2}}$$\displaystyle\Rightarrow -4=-\frac{1}{4}-\frac{5}{4}-D\Rightarrow {\color{pink}D=\frac{5}{2}}$

Let $x=2$${\color{brown}x=2}$. So, ${\displaystyle \Rightarrow 2^2+2(2)-4=-\frac{1}{4}(2+1)(2^2+1)+\frac{5}{4}(2-1)(2^2+1)+(C(2)+\frac{5}{2})(2-1)(2+1)}$$\displaystyle\Rightarrow {\color{brown}2}^2+2({\color{brown}2})-4={\color{orange}-\frac{1}{4}}({\color{brown}2}+1)({\color{brown}2}^2+1)+{\color{green}\frac{5}{4}}({\color{brown}2}-1)({\color{brown}2}^2+1)+(C{\color{brown}(2)}+{\color{pink}\frac{5}{2}})({\color{brown}2}-1)({\color{brown}2}+1)$

${\displaystyle \Rightarrow 4=-\frac{10}{4}+\frac{25}{4}+6C+\frac{15}{2}\Rightarrow 6C=-6\Rightarrow C=-1}$$\displaystyle\Rightarrow 4=-\frac{10}{4}+\frac{25}{4}+6C+\frac{15}{2}\Rightarrow 6C=-6\Rightarrow{\color{brown}{C=-1}}$

3. Thus,

${\displaystyle \frac{x^2+2x-4}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}=\frac{-1/4}{x-1}+\frac{5/4}{x+1}+\frac{-x+5/2}{x^2+1}}$$\displaystyle \frac{x^2+2x-4}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}=\frac{{\color{orange}-1/4}}{x-1}+\frac{{\color{green}5/4}}{x+1}+\frac{{\color{brown}-}x+{\color{pink}5/2}}{x^2+1}$

Check Your Work

Use Technology to Verify Partial Fraction Decomposition

NOTE: The use of technology in this activity is to verify, i.e., check, your partial fraction decomposition. You must show work on assessments to receive credit for all Calculus II work.

1
(* Express (x^2+2x-4)/(x^4-1) as a sum of partial fractions *)
2
Apart[(x^2+2x-4)/(x^4-1)]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_Code.html.

2. Apart[ ] has one argument:

   1. The function to decompose into sum of partial fractions.

3. For help on using the Apart[ ] function:

   1. In Mathematica, execute the code: $\text{?Apart}$$\text{?Apart}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the Apart[ ] function – you will be able to copy-paste code.

4. You may need parens, $($$(\,$ and $)$$\,)$, to group multiple terms in the numerator and denominator.

5. Remember, correct Mathematica code will be all black except for variables.

6. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Python

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Python syntax is located at: https://www.jjw3.com/Common_Python_Code.html.

2. All code is included above with comment codes that explain the executable code.

3. integrate( ) for indefinite integrals has two arguments:

   1. The integrand, i.e., function of the definite integral;

   2. $x$$x$​, where $x$$x$​​ is the variable used in the integrand.

4. You only need to change the function in the integrate( ) command.

5. The Python syntax for ${\cos }^2(x)$$\cos^2{(x)}$​​​ is $\text{cos(x)**2}$$\text{cos(x)**2}$​​​; etc.

6. Python requires $\ast$$*$ for explicit multiplication; $\ast \ast$$**$​ for exponents.

7. Remember to use parens $($$\left(\right.$ and $)$$\left.\right)$​ to group multiple terms in the numerator or denominator of a rational expression.

8. Python does not return the antiderivative with $+C$$+C$. You need to use $+C$$+C$ for all steps that are antiderivatives.

9. To execute code, press the “View the result” button:

Example 02: Express ${\displaystyle \frac{x+2}{x^4+2x^2}}$$\displaystyle \frac{x+2}{x^4+2x^2}$ as a sum of partial fractions.

Solution 02:

All three conditions needed to use the Method of Partial fractions are met.

Completely, factor the denominator: $x^4+2x^2=x^2(x^2+2)$$x^4+2x^2=x^2(x^2+2)$.

There are three (3) factors of the denominator; $x^2+2$$x^2+2$, $x$$x$, and another $x$$x$; thus, there must be three (3) distinct partial fractions.

Two of the factors in the denominator are linear factors, so a generic constant, $A$$A$ and $B$$B$, is used for the numerator: $\frac{A}{x}$$\frac{A}{x}$ and $\frac{B}{x^2}$$\frac{B}{x^2}$.

NOTE: $\frac{A}{x}$$\frac{A}{x}$ and $\frac{B}{x}$$\frac{B}{x}$ does NOT work since these two fractions can be combined into a single partial fraction $\frac{A+B}{x}$$\frac{A+B}{x}$ and the Method of Partial Fractions requires three (3) partial fractions for this rational expression.

The remaining factor of the denominator is an irreducible quadratic, so a generic linear function is used for the numerator: $\frac{Cx+D}{x^2+2}$$\frac{Cx+D}{x^2+2}$

Thus, the rational expression can be rewritten as the sum of three (3) partial fractions:

${\displaystyle \frac{x+2}{x^4+2x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+2}}$$\displaystyle \frac{x+2}{x^4+2x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+2}$

Next, we need to solve for the coefficients: $A$$A$, $B$$B$, $C$$C$, and $D$$D$.

1. Distribute the factors of the denominator to the entire equation and cancel factors:

${\displaystyle (\frac{x+2}{x^4+2x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+2})\left(x^2(x^2+2)\right)}$$\displaystyle \left(\frac{x+2}{x^4+2x^2}={\color{brown}\frac{A}{x}}+{\color{blue}\frac{B}{x^2}}+{\color{green}\frac{Cx+D}{x^2+2}}\right)\left({\color{red}x^2(x^2+2)}\right)$

Consider each product:

${\displaystyle \Rightarrow \frac{x+2}{\cancel{x^4+2x^2}}\cancel{x^2(x^2+2)}\Rightarrow x+2}$$\displaystyle \Rightarrow\frac{x+2}{\cancel{x^4+2x^2}}{\cancel{\color{red}x^2(x^2+2)}}\Rightarrow x+2$, and

${\displaystyle \Rightarrow \frac{A}{\cancel{x}}\cancel{x}x(x^2+2)\Rightarrow Ax(x^2+2)}$$\displaystyle \Rightarrow{\color{brown}\frac{A}{\cancel{x}}}{\color{red}\cancel{x}x(x^2+2)}\Rightarrow Ax(x^2+2)$, and

${\displaystyle \Rightarrow \frac{B}{\cancel{x^2}}\cancel{x^2}(x^2+2)\Rightarrow B(x^2+2)}$$\displaystyle \Rightarrow{\color{blue}\frac{B}{\cancel{x^2}}}{\color{red}\cancel{x^2}(x^2+2)}\Rightarrow B(x^2+2)$

${\displaystyle \Rightarrow \frac{Cx+D}{\cancel{x^2+2}}{x}^2\cancel{(x^2+2)}\Rightarrow (Cx+d)x^2}$$\displaystyle \Rightarrow{\color{green}\frac{Cx+D}{\cancel{x^2+2}}}{\color{red}x^2\cancel{(x^2+2)}}\Rightarrow (Cx+d)x^2$

Thus,

${\displaystyle \Rightarrow x+2=Ax(x^2+2)+B(x^2+2)+(Cx+D)x^2}$$\displaystyle\Rightarrow x+2=Ax(x^2+2)+B(x^2+2)+(Cx+D)x^2$

The traditional method suggests to expand, refactor in terms of powers of $x$$x$, set-up four (4) equations and use algebra to solve for the four (4) unknowns.

2. Alternate method. Do NOT expand.

Let $x=2$${\color{orange}x=2}$. So, ${\displaystyle \Rightarrow 0+2=A(0)(0^2+2)+B(0^2+2)+(C(0)+D)(0^2)}$$\displaystyle\Rightarrow {\color{orange}0}+2=A({\color{orange}0})({\color{orange}0}^2+2)+B({\color{orange}0}^2+2)+(C({\color{orange}0})+D)({\color{orange}0}^2)$

${\displaystyle \Rightarrow 2=B(2)\Rightarrow B=1}$$\displaystyle\Rightarrow 2=B(2)\Rightarrow {\color{orange}B=1}$

Let $x=1$${\color{green}x=1}$. So, ${\displaystyle \Rightarrow 1+2=A(1)(1^2+2)+(1)(1^2+2)+(C(1)+D)(1^2)\Rightarrow 3=3A+3+C+D}$$\displaystyle\Rightarrow {\color{green}1}+2=A{({\color{green}1})}({\color{green}1}^2+2)+{\color{orange}(1)}({\color{green}1}^2+2)+(C{\color{green}(1)}+D)({\color{green}1}^2)\Rightarrow 3=3A+3+C+D$

${\displaystyle \Rightarrow 0=3A+C+D}$$\displaystyle\Rightarrow 0=3A+C+D$ (equation 1)

Let $x=-1$${\color{pink}x=-1}$. So, ${\displaystyle \Rightarrow -1+2=A(-1)({(-1)}^2+2)+(1)({(-1)}^2+2)+(C(-1)+D)({(-1)}^2)\Rightarrow 1=-3A+3-C+D}$$\displaystyle\Rightarrow {\color{pink}-1}+2=A({\color{pink}-1})({\color{pink}(-1)}^2+2)+{\color{orange}(1)}({\color{pink}(-1)}^2+2)+(C{\color{pink}(-1)}+D)({\color{pink}(-1)}^2)\Rightarrow 1=-3A+3-C+D$

${\displaystyle \Rightarrow -2=-3A-C+D}$$\displaystyle\Rightarrow -2=-3A-C+D$ (equation 2)

Let $x=2$${\color{brown}x=2}$. So, ${\displaystyle \Rightarrow 2+2=A(2)(2^2+2)+(1)(2^2+2)+(C(2)+D)(2^2)\Rightarrow 4=12A+6+8C+4D}$$\displaystyle\Rightarrow {\color{brown}2}+2=A{\color{brown}(2)}({\color{brown}2}^2+2)+{\color{orange}(1)}({\color{brown}2}^2+2)+(C{\color{brown}(2)}+D)({\color{brown}2}^2)\Rightarrow 4=12A+6+8C+4D$

${\displaystyle \Rightarrow -2=12A+8C+4D}$$\displaystyle\Rightarrow -2=12A+8C+4D$ (equation 3)

We have three equations (labeled) with three unknowns ($A$$A$, $C$$C$, and $D$$D$). Either use Algebra to solve for the unknowns or use Mathematica:

Use Technology to Solve for Unknowns in a System of Equations

NOTE: You are permitted to use Mathematica to algebraically solve systems of equations without showing any work.

Mathematica

xxxxxxxxxx
2
1
(* Solve system of 3 equations with three unknowns in Example 02 *)
2
Solve[{0==3A+C+D, -2==-3A-C+D, -2==12A+8C+4D},{A,C,D}]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_Code.html.

2. Solve[ ] has at least two arguments:

   1. The equations to be solved; the equations are grouped between $\{$$\{$ and $\}$$\}$ and are separated by commas

   2. The variables/parameters to be solved; the parameters are grouped between $\{$$\{$ and $\}$$\}$ and are separated by commas

3. For help on using the Solve[ ] function:

   1. In Mathematica, execute the code: $\text{?Solve}$$\text{?Solve}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the Solve[ ] function – you will be able to copy-paste code.

4. Mathematica uses the double equals sign, $==$$==$, to represent the equals sign in an equation.

5. Remember, correct Mathematica code will be all black except for variables.

6. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Investigation 09

Rewrite the following rational expressions into a sum of partial fractions, if possible:

1. ${\displaystyle \frac{1}{x^2-3x-4}}$$\displaystyle \frac{1}{x^2-3x-4}$

2. ${\displaystyle \frac{x}{x^2-5x+4}}$$\displaystyle \frac{x}{x^2-5x+4}$

3. ${\displaystyle \frac{x-2}{x^2-9}}$$\displaystyle \frac{x-2}{x^2-9}$

4. ${\displaystyle \frac{3x+2}{x^2-x}}$$\displaystyle \frac{3x+2}{x^2-x}$

5. ${\displaystyle \frac{x+1}{x^2-4}}$$\displaystyle \frac{x+1}{x^2-4}$

6. ${\displaystyle \frac{3}{x^3+x}}$$\displaystyle \frac{3}{x^3+x}$

7. ${\displaystyle \frac{2-x}{x^3-1}}$$\displaystyle \frac{2-x}{x^3-1}$

8. ${\displaystyle \frac{x}{x^3+8}}$$\displaystyle \frac{x}{x^3+8}$

9. ${\displaystyle \frac{x^2+x+1}{x^3-2x^2+x-2}}$$\displaystyle \frac{x^2+x+1}{x^3-2x^2+x-2}$

10. ${\displaystyle \frac{x-3}{x^4-16}}$$\displaystyle \frac{x-3}{x^4-16}$

11. ${\displaystyle \frac{x^2+2}{x^5-3x^4+2x^3}}$$\displaystyle \frac{x^2+2}{x^5-3x^4+2x^3}$

Integrating Rational Functions

Example 03: Evaluate ${\displaystyle \int \frac{x^2+2x-4}{x^4-1}dx}$$\displaystyle \int{\frac{x^2+2x-4}{x^4-1}\,dx}$.

Solution 03: From Example 01, we know

${\displaystyle \int \frac{x^2+2x-4}{x^4-1}dx=\int (\frac{-1/4}{x-1}+\frac{5/4}{x+1}+\frac{-x+5/2}{x^2+1})dx}$$\displaystyle \int{\frac{x^2+2x-4}{x^4-1}\,dx}=\int{\left(\frac{-1/4}{x-1}+\frac{5/4}{x+1}+\frac{-x+5/2}{x^2+1}\right)\,dx}$

${\displaystyle \Rightarrow -\frac{1}{4}\int \frac{1}{x-1}dx+\frac{5}{4}\int \frac{1}{x+1}dx-\int \frac{x}{x^2+1}dx+\frac{5}{2}\int \frac{1}{x^2+1}dx}$$\displaystyle \Rightarrow -\frac{1}{4}\int{\frac{1}{x-1}\,dx} + \frac{5}{4}\int{\frac{1}{x+1}\,dx} - \int{\frac{x}{x^2+1}\,dx} + \frac{5}{2}\int{\frac{1}{x^2+1}\,dx}$

Using $u$$u$-Substitution (for the third term) and known integrals in for the remaining three terms,

${\displaystyle \Rightarrow \int \frac{x^2+2x-4}{x^4-1}dx=-\frac{1}{4}\ln |x-1|+\frac{5}{4}\ln |x+1|-\frac{1}{2}\ln |x^2+1|+\frac{5}{2}{\tan }^{-1}(x)+C}$$\displaystyle\Rightarrow \int{\frac{x^2+2x-4}{x^4-1}\,dx}=-\frac{1}{4}\ln{|\,x- 1\,|} + \frac{5}{4}\ln{|\,x+1\,|} - \frac{1}{2}\ln{|\,x^2+1\,|} + \frac{5}{2}\tan^{-1}{(x)}+C$

Check Your Work

Use Technology to Verify Antiderivatives

NOTE: The use of technology in this activity is to verify, i.e., check, your answers. You must show work on assessments to receive credit for all Calculus II work.

Mathematica

xxxxxxxxxx
2
1
(* Integrate (x^2+2x-4)/(x^4-1) *)
2
Integrate[(x^2+2x-4)/(x^4-1), x]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_Code.html.

2. Integrate[ ] has two arguments:

   1. The function in the integrand;

   2. The independent variable of the integral.

3. For help on using the Integrate[ ] function:

   1. In Mathematica, execute the code: $\text{?Integrate}$$\text{?Integrate}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the Integrate[ ] function – you will be able to copy-paste code.

4. You may need parens, $($$(\,$ and $)$$\,)$, to group multiple terms in the numerator and denominator.

5. Remember, correct Mathematica code will be all black except for variables.

6. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

7. Mathematica does not show $+C$$+C$ for antiderivatives. You will need to remember to include $+C$$+C$ in your answers.

8. Mathematica may return a result in a different, but equivalent, form than your answer. In these cases, use FTC-I to check your work.

Investigation 10

Integrate the following rational functions.

1. ${\displaystyle \int \frac{1}{x^2-3x-4}\text{d}x}$$\displaystyle \int{\frac{1}{x^2-3x-4}\,\text{d}x}$

2. ${\displaystyle \int \frac{x}{x^2-5x+4}\text{d}x}$$\displaystyle \int{\frac{x}{x^2-5x+4}\,\text{d}x}$

3. ${\displaystyle \int \frac{x-2}{x^2-9}\text{d}x}$$\displaystyle \int{\frac{x-2}{x^2-9}\,\text{d}x}$

4. ${\displaystyle \int \frac{3x+2}{x^2-x}\text{d}x}$$\displaystyle \int{\frac{3x+2}{x^2-x}\,\text{d}x}$

5. ${\displaystyle \int \frac{x+1}{x^2-4}\text{d}x}$$\displaystyle \int{\frac{x+1}{x^2-4}\,\text{d}x}$

6. ${\displaystyle \int \frac{3}{x^3+x}\text{d}x}$$\displaystyle \int{\frac{3}{x^3+x}\,\text{d}x}$

7. ${\displaystyle \int \frac{2-x}{x^3-1}\text{d}x}$$\displaystyle \int{\frac{2-x}{x^3-1}\,\text{d}x}$

8. ${\displaystyle \int \frac{x}{x^3+8}\text{d}x}$$\displaystyle \int{\frac{x}{x^3+8}\,\text{d}x}$

9. ${\displaystyle \int \frac{x^2+x+1}{x^3-2x^2+x-2}\text{d}x}$$\displaystyle \int{\frac{x^2+x+1}{x^3-2x^2+x-2}\,\text{d}x}$

10. ${\displaystyle \int \frac{x-3}{x^4-16}\text{d}x}$$\displaystyle \int{\frac{x-3}{x^4-16}\,\text{d}x}$

11. ${\displaystyle \int \frac{x^2+2}{x^5-3x^4+2x^3}\text{d}x}$$\displaystyle \int{\frac{x^2+2}{x^5-3x^4+2x^3}\,\text{d}x}$

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Created: Monday, 6 September 2020 13:33 EDT Last Modified: Wednesday, 17 August 2022 - 09:49 (EDT)