Lila: /* Resources */

2021-10-29T17:09:37Z

Resources

Lila: /* Exercises */

2021-10-06T23:03:42Z

Exercises

Lila: /* Exercises */

2021-10-06T23:03:10Z

Exercises

Lila: /* Exercise */

2021-10-06T23:02:47Z

Exercise

Lila at 23:02, 6 October 2021

2021-10-06T23:02:07Z

James.kercheville: Added video links

2020-08-24T18:14:51Z

Added video links

New page

[https://youtu.be/8MMALX7mUqQ Integration Using Partial Fraction Decomposition Part 1] by James Sousa

[https://youtu.be/BgVNLWLJmMs Integration Using Partial Fraction Decomposition Part 2] by James Sousa

[https://youtu.be/c2oLHtPA03U Partial Fraction Decomposition Part 1 (Linear)] by James Sousa

[https://youtu.be/SQ4l8yDlBVE Partial Fraction Decomposition - Part 2 of 2] by James Sousa

[https://youtu.be/HZTv4zCgEnA Partial Fraction Decomposition - Example 1] by patrickJMT

[https://youtu.be/hnIs8uwAT3U Partial Fraction Decomposition - Example 2] by patrickJMT

[https://youtu.be/lH9Mw65vAIA Partial Fraction Decomposition - Example 4] by patrickJMT

[https://youtu.be/Slfrb8aEj08 Partial Fraction Decomposition - Example 5] by patrickJMT

[https://youtu.be/_YMG5mQTAOg Partial Fraction Decomposition - Example 6] by patrickJMT

[https://youtu.be/7GcSOjLitF8 Partial Fraction Decompositions] by patrickJMT

← Older revision		Revision as of 17:09, 29 October 2021
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	[https://youtu.be/7GcSOjLitF8 Partial Fraction Decompositions] by patrickJMT		[https://youtu.be/7GcSOjLitF8 Partial Fraction Decompositions] by patrickJMT
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikibooks.org/wiki/Calculus/Integration_techniques/Partial_Fraction_Decomposition Partial Fraction Decomposition, Wikibooks: Calculus/Integration techniques] under a CC BY-SA license

@@ Line 78: / Line 78: @@
 ===Exercises===
 Evaluate the following by the method partial fraction decomposition.
 . <math>\int\frac{2x+11}{(x+6)(x+5)}dx</math>

@@ Line 119: / Line 119: @@
 ====Exercises====
 Evaluate the following using the method of partial fractions.
 . <math>\int\frac{2}{(x+2)(x^2+3)}dx</math>

@@ Line 129: / Line 129: @@
 ====Exercise====
 Evaluate the following using the method of partial fractions.
 . <math>\int\frac{dx}{(x-1)(x^2+1)^2}</math>

← Older revision		Revision as of 23:02, 6 October 2021
Line 1:		Line 1:
		+	Suppose we want to find <math>\int\frac{3x+1}{x^2+x}dx</math> . One way to do this is to simplify the integrand by finding constants <math>A</math> and <math>B</math> so that
		+	:<math>\frac{3x+1}{x^2+x}=\frac{3x+1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}</math> .
		+	This can be done by cross multiplying the fraction which gives
		+	:<math>\frac{3x+1}{x(x+1)}=\frac{A(x+1)+Bx}{x(x+1)}</math>
		+	As both sides have the same denominator we must have
		+	:<math>3x+1=A(x+1)+Bx</math>
		+	This is an equation for <math>x</math> so it must hold whatever value <math>x</math> is. If we put in <math>x=0</math> we get
		+	<math>A=1</math>
		+	and putting <math>x=-1</math> gives <math>=-B=-2</math> so <math>B=2</math> .
		+	So we see that
		+	:<math>\frac{3x+1}{x^2+x}=\frac{1}{x}+\frac{2}{x+1}</math>
		+
		+	Returning to the original integral
		+	:{\|
		+	\|<math>\int\frac{3x+1}{x^2+x}dx</math>
		+	\|<math>=\int\frac{dx}{x}+\int\frac{2}{x+1}dx</math>
		+	\|-
		+	\|
		+	\|<math>=\int\frac{dx}{x}+2\int\frac{dx}{x+1}</math>
		+	\|-
		+	\|
		+	\|<math>=\ln\|x\|+2\ln\Big\|x+1\Big\|+C</math>
		+	\|}
		+
		+	Rewriting the integrand as a sum of simpler fractions has allowed us to reduce the initial integral to a sum of simpler integrals. In fact this method works to integrate any rational function.
		+
		+	==Method of Partial Fractions==
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	To decompose the rational function <math>\frac{P(x)}{Q(x)}</math>:
		+	*'''Step 1''' Use long division (if necessary) to ensure that the degree of <math>P(x)</math> is less than the degree of <math>Q(x)</math>.
		+	*'''Step 2''' Factor Q(x) as far as possible.
		+	*'''Step 3''' Write down the correct form for the partial fraction decomposition (see below) and solve for the constants.
		+	</blockquote>
		+
		+	To factor Q(x) we have to write it as a product of linear factors (of the form <math>ax+b</math>) and irreducible quadratic factors (of the form <math>ax^2+bx+c</math> with <math>b^2-4ac<0</math>).
		+
		+	Some of the factors could be repeated. For instance if <math>Q(x) = x^3-6x^2+9x</math> we factor <math>Q(x)</math> as
		+	:<math>Q(x)=x(x^2-6x+9)=x(x-3)(x-3)=x(x-3)^2</math>
		+
		+	It is important that in each quadratic factor we have <math>b^2-4ac<0</math> , otherwise it is possible to factor that quadratic piece further. For example if <math>Q(x)=x^3-3x^2+2x</math> then we can write
		+	:<math>Q(x)=x(x^2-3x+2)=x(x-1)(x-2)</math>
		+
		+	We will now show how to write <math>\frac{P(x)}{Q(x)}</math> as a sum of terms of the form
		+	:<math>\frac{A}{(ax+b)^k}</math> and <math>\frac{Ax+B}{(ax^2+bx+c)^k}</math>
		+	Exactly how to do this depends on the factorization of <math>Q(x)</math> and we now give four cases that can occur.
		+
		+	===Q(x) is a product of linear factors with no repeats===
		+	This means that <math>Q(x)=(a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)</math> where no factor is repeated and no factor is a multiple of another.
		+
		+	For each linear term we write down something of the form <math>\frac{A}{(ax+b)}</math> , so in total we write
		+
		+	:<math>\frac{P(x)}{Q(x)}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{a_2x+b_2}+\cdots+\frac{A_n}{a_nx+b_n}</math>
		+
		+	====Example Problem====
		+	Find <math>\int\frac{1+x^2}{(x+3)(x+5)(x+7)}dx</math>
		+
		+	Here we have <math>P(x)=1+x^2\ ,\ Q(x)=(x+3)(x+5)(x+7)</math> and ''Q(x)'' is a product of linear factors. So we write
		+
		+	<math>\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{A}{x+3}+\frac{B}{x+5}+\frac{C}{x+7}</math>
		+
		+	Multiply both sides by the denominator
		+
		+	<math>1+x^2=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)</math>
		+
		+	Substitute in three values of ''x'' to get three equations for the unknown constants,
		+
		+	<math>\begin{matrix} x=-3 & 1+3^2=2\cdot 4 A \\ x=-5 & 1+5^2=-2\cdot 2 B \\
		+	x=-7 & 1+7^2=(-4)\cdot (-2)C \end{matrix}</math>
		+
		+	so <math>A=\tfrac{5}{4}\ ,\ B=-\tfrac{13}{2}\ ,\ C=\tfrac{25}{4}</math> , and
		+
		+	<math>\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{5}{4x+12}-\frac{13}{2x+10}+\frac{25}{4x+28}</math>
		+
		+	We can now integrate the left hand side.
		+
		+	<math>\int\frac{1+x^2}{(x+3)(x+5)(x+7)}dx=\tfrac{5}{4}\ln\Big\|x+3\Big\|-\tfrac{13}{2}\ln\Big\|x+5\Big\|+\tfrac{25}{4}\ln\Big\|x+7\Big\|+C</math>
		+
		+	===Exercises===
		+	Evaluate the following by the method partial fraction decomposition.
		+	1. <math>\int\frac{2x+11}{(x+6)(x+5)}dx</math>
		+
		+	2. <math>\int\frac{7x^2-5x+6}{(x-1)(x-3)(x-7)}dx</math>
		+
		+	===Q(x) is a product of linear factors some of which are repeated===
		+	If <math>(ax+b)</math> appears in the factorisation of <math>Q(x)</math> ''k''-times then instead of writing the piece <math>\frac{A}{ax+b}</math> we use the more complicated expression
		+
		+	<math>\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\frac{A_3}{(ax+b)^3}+\cdots+\frac{A_k}{(ax+b)^k}</math>
		+
		+	====Example 2====
		+	Find <math>\int\frac{dx}{(x+1)(x+2)^2}</math>
		+
		+	Here <math>P(x)=1</math> and <math>Q(x)=(x+1)(x+2)^2</math> We write
		+
		+	<math>\frac{1}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}</math>
		+
		+	Multiply both sides by the denominator
		+	<math>1=A(x+2)^2+B(x+1)(x+2)+C(x+1)</math>
		+
		+	Substitute in three values of <math>x</math> to get 3 equations for the unknown constants,
		+
		+	<math>\begin{matrix} x=0 & 1=2^2A+2B+C \\ x=-1 & 1=A \\
		+	x=-2 & 1=-C \end{matrix}</math>
		+
		+	so <math>A=1\ ,\ B=-1\ ,\ C=-1</math> and
		+
		+	<math>\frac{1}{(x+1)(x+2)^2}=\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{(x+2)^2}</math>
		+
		+	We can now integrate the left hand side.
		+	:<math>\int\frac{dx}{(x+1)(x+2)^2}=\ln\left\|x+1\right\|-\ln\left\|x+2\right\|+\frac{1}{x+2}+C</math>
		+	We now simplify the fuction with the property of Logarithms.
		+	:<math>\ln\left\|x+1\right\|-\ln\left\|x+2\right\|+\frac{1}{x+2}+C=\ln\left\|\frac{x+1}{x+2}\right\|+\frac{1}{x+2}+C</math>
		+
		+	====Exercise====
		+	3. Evaluate <math>\int\frac{x^2-x+2}{x(x+2)^2}dx</math> using the method of partial fractions.
		+
		+	===Q(x) contains some quadratic pieces which are not repeated===
		+	If <math>ax^2+bx+c</math> appears we use <math>\frac{Ax+B}{ax^2+bx+c}</math> .
		+
		+	====Exercises====
		+	Evaluate the following using the method of partial fractions.
		+	4. <math>\int\frac{2}{(x+2)(x^2+3)}dx</math>
		+
		+	5. <math>\int\frac{dx}{(x+2)(x^2+2)}</math>
		+
		+	===Q(x) contains some repeated quadratic factors===
		+	If <math>ax^2+bx+c</math> appears k-times then use
		+	:<math>\frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+\frac{A_3x+B_3}{(ax^2+bx+c)^3}+\cdots+\frac{A_kx+B_k}{(ax^2+bx+c)^k}</math>
		+
		+	====Exercise====
		+	Evaluate the following using the method of partial fractions.
		+	6. <math>\int\frac{dx}{(x-1)(x^2+1)^2}</math>
		+
		+	===Exercies Solutions===
		+	# <math>\ln\Big\|x+6\Big\|+\ln\Big\|x+5\Big\|+C</math>
		+	# <math>\tfrac{2}{3}\ln\Big\|x-1\Big\|-\tfrac{27}{4}\ln\Big\|x-3\Big\|+\tfrac{157}{12}\ln\Big\|x-7\Big\|+C</math>
		+	# <math>\frac{\ln\Big\|x(x+2)\Big\|}{2}+\frac{4}{x+2}+C</math>
		+	# <math>\ln\left(\sqrt[7]{\frac{(x+2)^2}{x^2+3}}\right)+\frac{4\arctan\Big(\tfrac{x}{\sqrt3}\Big)}{7\sqrt3}+C</math>
		+	# <math>\ln\left(\sqrt[12]{\frac{(x+2)^2}{x^2+2}}\right)+\frac{\sqrt2\arctan\Big(\tfrac{x}{\sqrt2}\Big)}{6}+C</math>
		+	# <math>\frac{1-x}{4(x^2+1)}+\tfrac{1}{8}\ln\left(\frac{(x-1)^2}{x^2+1}\right)-\frac{\arctan(x)}{2}+C</math>
		+
		+	==Resources==
	[https://youtu.be/8MMALX7mUqQ Integration Using Partial Fraction Decomposition Part 1] by James Sousa		[https://youtu.be/8MMALX7mUqQ Integration Using Partial Fraction Decomposition Part 1] by James Sousa

Partial Fractions - Revision history

Lila: /* Resources */

Lila: /* Exercises */

Lila: /* Exercises */

Lila: /* Exercise */

Lila at 23:02, 6 October 2021

James.kercheville: Added video links