Simon's copy of Integration by partial fractions. (Video)

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Simon Thomas 6 years ago

Created this as a copy of Integration by partial fractions. (Video).

Name	Status	Author	Last Modified
Integration by partial fractions. (Video)	Ready to use	Bill Foster	01/06/2016 09:47
Numbas demo: video	Ready to use	Christian Lawson-Perfect	25/08/2020 15:00
YJ's copy of Numbas demo: video	Ready to use	YJ Kim	01/06/2016 09:47
Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 14:06
Simon's copy of Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 14:09
Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 15:18
Simon's copy of Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 15:31
Simon's copy of Simon's copy of Simon's copy of Integration by partial fractions. (Video)	draft	Simon Thomas	06/03/2019 15:33

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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	9
d	integer	2
d1	integer	2
s2	integer	1
s3	integer	1
b1	integer	4
s1	integer	1
b	integer	4
c	integer	4

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Value

xxxxxxxxxx
 
s1*random(1..9)

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Generated value: `integer`

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Parts

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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.

There is a video in Show steps which solves a similar example.

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

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

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

One way to find A and B is to compare coefficients:

Identifying coefficients:

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

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

On solving these simultaneous 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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Simon's copy of Integration by partial fractions. (Video)

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

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Simon Thomas 6 years ago

Error in variable testing condition

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Generated value: `integer`

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Simon's copy of Integration by partial fractions. (Video)

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`