Integration by partial fractions

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Christian Lawson-Perfect

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Christian Lawson-Perfect 10 years 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	23/09/2020 20:02
Simon's copy of Integration by partial fractions	draft	Simon Thomas	06/03/2019 14:06
Simon's copy of 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 15:14

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Question statement

Find the following integral.

$I = \simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}$

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	-6
c	integer	2
b	integer	-7
s3	integer	-1
s2	integer	-1
s1	integer	-1
b1	integer	-7

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s1*random(1..9)

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Generated value: `integer`

-6

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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

First we factorise $\simplify[std]{x^2+{a+b}*x+{a*b}=(x+{a})*(x+{b})}$ .

You can do this by spotting the factors or by completing the square.

Next we use partial fractions 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:

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

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:

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

$\begin{eqnarray*} I &=& \simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}\\ &=& \simplify[std]{Int({c}/((x +{a})*(x+{b})),x )}\\ &=& \simplify[std]{({c}/{b-a})*(Int(1/(x+{a}),x) -Int(1/(x+{b}),x))}\\ &=& \simplify[std]{({c}/{b-a})*(ln(x+{a})-ln(x+{b}))+C} \end{eqnarray*}$

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Taxonomy: mathcentre

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Christian Lawson-Perfect 10 years 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 10 years ago

Bill Foster 55 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`