// Numbas version: finer_feedback_settings {"name": "Simon's copy of Integration by partial fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"statement": "\n

Find the following integral.

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

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Input all numbers as fractions or integers and not decimals.

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Input the constant of integration as $C$.

\n \n ", "type": "question", "name": "Simon's copy of Integration by partial fractions", "variable_groups": [], "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "questions": [], "pickQuestions": 0}], "progress": "ready", "extensions": [], "parts": [{"marks": 0.0, "prompt": "\n

$I=\\;$[[0]]

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Input all numbers as fractions or integers and not decimals.

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Input the constant of integration as $C$.

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

\n \n ", "steps": [{"marks": 0.0, "type": "information", "prompt": "\n

Use partial fractions in order to write:
\\[\\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\\; = \\simplify[std]{A/(x+{a})+B/(x+{b})}\\]

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for suitable integers or fractions $A$ and $B$.

\n "}], "type": "gapfill", "stepspenalty": 1.0, "gaps": [{"vsetrangepoints": 5.0, "marks": 3.0, "checkingtype": "absdiff", "answer": "({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C", "vsetrange": [11.0, 12.0], "checkingaccuracy": 0.0001, "type": "jme", "answersimplification": "std", "notallowed": {"message": "

Input all numbers as fractions or integers and not decimals.

", "strings": ["."], "partialcredit": 0.0, "showstrings": false}}]}], "advice": "\n

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:

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

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Identifying coefficients:

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

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

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

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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}))}\\]

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So \\[\\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*}\\]

\n ", "functions": {}, "variables": {"s2": {"name": "s2", "definition": "random(1,-1)"}, "c": {"name": "c", "definition": "random(2..9)"}, "d": {"name": "d", "definition": "if(d1=a*c,if(d1+1=b*c,d1+2,d1+1),if(d1=b*c,if(d1+1=a*c,d1+2,d1+1),d1))"}, "s1": {"name": "s1", "definition": "random(1,-1)"}, "a": {"name": "a", "definition": "s1*random(1..9)"}, "s3": {"name": "s3", "definition": "random(1,-1)"}, "b": {"name": "b", "definition": "if(b1=a,b1+s3,b1)"}, "b1": {"name": "b1", "definition": "s2*random(1..9)"}, "d1": {"name": "d1", "definition": "s3*random(1..9)"}}, "tags": ["2 distinct linear factors", "Calculus", "Steps", "calculus", "compare coefficients", "identify coefficients", "integrals", "integration", "logarithms", "partial fractions", "steps", "two distinct linear factors"], "showQuestionGroupNames": false, "metadata": {"notes": "\n \t\t \t\t

5/08/2012:

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Added tags.

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Added description.

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Added decimal point as forbidden string.

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Note the checking range is chosen so that the arguments of the log terms are always positive - could have used abs - might be better?

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Improved display of Advice. 

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Added information about Show steps, also introduced penalty of 1 mark.

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Added !noLeadingMinus to ruleset std for display purposes.

\n \t\t \n \t\t", "description": "

Find $\\displaystyle\\int \\frac{ax+b}{(x+c)(x+d)}\\;dx,\\;a\\neq 0,\\;c \\neq d $.

", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}