// Numbas version: exam_results_page_options {"name": "Maria'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": [{"type": "question", "parts": [{"scripts": {}, "type": "gapfill", "gaps": [{"answer": "({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C", "type": "jme", "answersimplification": "std", "showCorrectAnswer": true, "vsetrangepoints": 5, "expectedvariablenames": [], "vsetrange": [11, 12], "checkingtype": "absdiff", "scripts": {}, "notallowed": {"showStrings": false, "message": "
Input all numbers as fractions or integers and not decimals.
", "strings": ["."], "partialCredit": 0}, "marks": 3, "showpreview": true, "checkvariablenames": false, "checkingaccuracy": 0.0001}], "showCorrectAnswer": true, "stepsPenalty": 1, "marks": 0, "prompt": "\n$I=\\;$[[0]]
\nInput all numbers as fractions or integers and not decimals.
\nInput the constant of integration as $C$.
\nClick on Show steps for help if you need it. You will lose 1 mark if you do so.
\n \n ", "steps": [{"scripts": {}, "type": "information", "prompt": "\nUse partial fractions in order to write:
\\[\\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\\; = \\simplify[std]{A/(x+{a})+B/(x+{b})}\\]
for suitable integers or fractions $A$ and $B$.
\n ", "showCorrectAnswer": true, "marks": 0}]}], "question_groups": [{"pickingStrategy": "all-ordered", "name": "", "questions": [], "pickQuestions": 0}], "functions": {}, "tags": ["2 distinct linear factors", "Calculus", "calculus", "checked2015", "compare coefficients", "identify coefficients", "integration", "logarithms", "MAS1601", "mas1601", "partial fractions", "Steps", "steps", "two distinct linear factors"], "advice": "\nUsing 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:
$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}*x+{d}} \\Rightarrow \\simplify[std]{(A+B)*x+{b}*A+{a}*B={c}*x+{d}}$
\nIdentifying coefficients:
\nConstant term: $\\simplify[std]{{b}*A+{a}*B = {d}}$
\nCoefficent $x$: $ \\simplify[std]{A+B={c}}$ which gives $A =\\var{c} -B$
\nOn solving these equations we obtain $\\displaystyle \\simplify[std]{A = {d-a*c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={d-b*c}/{a-b}}$
\nWhich 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}))}\\]
\nSo \\[\\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 ", "showQuestionGroupNames": false, "name": "Maria's copy of Integration by partial fractions", "statement": "\nFind the following integral.
\n\\[I = \\simplify[std]{Int(({c}*x+{d})/((x +{a})*(x+{b})),x )}\\]
\nInput all numbers as fractions or integers and not decimals.
\nInput the constant of integration as $C$.
\n \n ", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "metadata": {"notes": "\n \t\t \t\t5/08/2012:
\n \t\t \t\tAdded tags.
\n \t\t \t\tAdded description.
\n \t\t \t\tAdded decimal point as forbidden string.
\n \t\t \t\tNote the checking range is chosen so that the arguments of the log terms are always positive - could have used abs - might be better?
\n \t\t \t\tImproved display of Advice.
\n \t\t \t\tAdded information about Show steps, also introduced penalty of 1 mark.
\n \t\t \t\tAdded !noLeadingMinus to ruleset std for display purposes.
\n \t\t \n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Find $\\displaystyle\\int \\frac{ax+b}{(x+c)(x+d)}\\;dx,\\;a\\neq 0,\\;c \\neq d $.
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