Luis's copy of Integration by partial fractions

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Luis Hernandez 6 years, 4 months ago

Created this as a copy of Integration by partial fractions.

Name	Status	Author	Last Modified
Integration by partial fractions	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Luis's copy of Integration by partial fractions	draft	Luis Hernandez	30/11/2018 18:33
Maria's copy of Integration by partial fractions	draft	Maria Aneiros	27/05/2019 07:25
Integration - By Partial Fractions	draft	Kevin Bohan	19/09/2019 11:55

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Question statement

Find the following integral.

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

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	-5
c	integer	5
b	integer	-7
d	integer	3
s3	integer	1
s2	integer	-1
s1	integer	-1
b1	integer	-7
d1	integer	3

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xxxxxxxxxx
 
s1*random(1..9)

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Generated value: `integer`

-5

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$I=\;$ Unnamed gap

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

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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:

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

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

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

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:

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

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

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Luis's copy of Integration by partial fractions

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Taxonomy: mathcentre

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Luis Hernandez 6 years, 4 months ago

Error in variable testing condition

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Generated value: `integer`

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Luis's copy of Integration by partial fractions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Luis Hernandez 6 years, 4 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`