Integration by partial fractions

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Name	Status	Author	Last Modified
Integration by partial fractions	Ready to use	Bill Foster	01/06/2016 09:47
CF Maths In class test three question 8 Integration by partial fractions with limits	Ready to use	Jinhua Mathias	28/08/2024 12:15
Andrew's copy of CF Maths In class test three question 8 Integration by partial fractions with limits	Ready to use	Andrew Dunbar	01/06/2016 13:58
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 14:06
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 14:07
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 15:10

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Question statement

Find the following integral.

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

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	-1
c	integer	4
b	integer	7
d	integer	-9
s3	integer	-1
s2	integer	1
s1	integer	-1
b1	integer	7
d1	integer	-9

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Description

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Generated value: `integer`

-1

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

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1. Gap Mathematical expression
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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:

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

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

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

Identifying coefficients:

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

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

On solving these equations we obtain $\displaystyle \simplify[std]{A = {d-a*c}/{b-a}}$ and $\displaystyle \simplify[std]{B={d-b*c}/{a-b}}$

Which gives:

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

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

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Christian Lawson-Perfect 9 years, 11 months ago

Bill Foster 55 years, 2 months ago

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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

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Generated value: integer

Parts

Gap-fill

Generated value: `integer`