First of all factorise the denominator.
\nYou have to find $a$ and $b$ such that $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+a)*(x+b)}$
\nThen use partial fractions to write:
\\[\\simplify[std]{({c}*x+{d})/((x +a)*(x+b)) = A/(x+a)+B/(x+b)}\\]
for suitable integers or fractions $A$ and $B$.
\n ", "extendBaseMarkingAlgorithm": true, "variableReplacements": []}], "showCorrectAnswer": true, "customMarkingAlgorithm": "", "adaptiveMarkingPenalty": 0, "showFeedbackIcon": true, "scripts": {}, "stepsPenalty": 1, "marks": 0, "gaps": [{"unitTests": [], "showCorrectAnswer": true, "variableReplacements": [], "showFeedbackIcon": true, "marks": 3, "useCustomName": false, "showPreview": true, "checkVariableNames": false, "extendBaseMarkingAlgorithm": true, "failureRate": 1, "vsetRangePoints": 5, "customName": "", "checkingAccuracy": 0.001, "adaptiveMarkingPenalty": 0, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "scripts": {}, "answerSimplification": "std", "type": "jme", "notallowed": {"message": "Input all numbers as fractions or integers and not decimals.
", "strings": ["."], "partialCredit": 0, "showStrings": false}, "vsetRange": [11, 12], "answer": "({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C", "checkingType": "absdiff", "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "useCustomName": false, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "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 ", "extendBaseMarkingAlgorithm": true, "sortAnswers": false, "variableReplacements": []}], "variable_groups": [], "statement": "\nFind the following integral.
\n\\[I = \\simplify[std]{Int(({c}*x+{d})/(x^2+{a+b}*x+{a*b}),x )}\\]
\nInput all numbers as fractions or integers and not decimals.
\nInput the constant of integration as $C$.
\n ", "extensions": [], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Factorise $x^2+cx+d$ into 2 distinct linear factors and then find $\\displaystyle \\int \\frac{ax+b}{x^2+cx+d}\\;dx,\\;a \\neq 0$ using partial fractions or otherwise.
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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:
$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}*x+{d}} \\Rightarrow \\simplify[std]{(A+B)*x+{b}*A+{a}*B={c}*x+{d}}$
\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})) = ({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^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*}\\]
", "ungrouped_variables": ["a", "c", "b", "d", "s3", "s2", "s1", "b1", "d1"], "functions": {}, "tags": [], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Kevin Bohan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3363/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Kevin Bohan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3363/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}