Integration by partial fractions

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Christian Lawson-Perfect

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Christian Lawson-Perfect 9 years, 11 months ago

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Bill Foster 55 years, 2 months ago

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Name	Status	Author	Last Modified
Integration by partial fractions	Ready to use	Bill Foster	01/06/2016 09:47
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 15:05

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Question statement

Find the following integral.

$I = \simplify[std]{Int({c}/((x +{a})*(x+{b})),x )}$

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	-8
c	integer	3
b	integer	-7
s3	integer	-1
s2	integer	-1
s1	integer	-1
b1	integer	-7
c1	integer	3

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s1*random(1..9)

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Description

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Generated value: `integer`

-8

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Statement
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"Unnamed part" → "Unnamed gap" - Correct answer
"Unnamed part" → "Unnamed step" - prompt

Parts

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1. Gap Mathematical expression
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1. Step Information only

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$I=\;$ Unnamed gap

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

Click on Show steps for help if you need it. You will lose 1 mark if you do so.

Advice

Using partial fractions we have to find $A$ and $B$ such that:

$\simplify[std]{{c}/((x +{a})*(x+{b})) = A/(x+{a})+B/(x+{b})}$
Multiplying both sides of the equation by

$\displaystyle \simplify[std]{1/((x +{a})*(x+{b}))}$ we obtain:

$\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}}\Rightarrow \simplify[std]{(A+B)*x+{b}A+{a}B={c}}$ .

We now identify coefficients on both sides of this equation.

Identifying coefficients:

Constant term: $\simplify[std]{{b}*A+{a}*B = {c}}$

Coefficent $x$ : $\simplify[std]{A + B = 0}$ which gives $A = -B$

Hence we obtain $\displaystyle \simplify[std]{A = {c}/{b-a}}$ and $\displaystyle \simplify[std]{B={-c}/{b-a}}$

Which gives:

$\simplify[std]{{c}/((x +{a})*(x+{b})) = ({c}/{b-a})*(1/(x+{a}) -1/(x+{b}))}$

$\begin{eqnarray*} I &=& \simplify[std]{Int({c}/((x +{a})*(x+{b})),x )}\\ &=& \simplify[std]{({c}/{b-a})*(Int(1/(x+{a}),x) -Int(1/(x+{b}),x))}\\ &=& \simplify[std]{({c}/{b-a})*(ln(x+{a})-ln(x+{b}))+C} \end{eqnarray*}$

Or equivalently: $\displaystyle I= \simplify[std]{({c}/{b-a})*(ln((x+{a})/(x+{b})))+C}$

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Taxonomy: mathcentre

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Christian Lawson-Perfect 9 years, 11 months ago

Bill Foster 55 years, 2 months ago

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Generated value: `integer`

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Integration by partial fractions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 9 years, 11 months ago

Bill Foster 55 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`