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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:
$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}*x+{d}} \\Rightarrow \\simplify[std]{(A+B)*x+{b}*A+{a}*B={c}*x+{d}}$
\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*}\\]
", "preamble": {"css": "", "js": ""}, "question_groups": [{"questions": [], "pickingStrategy": "all-ordered", "pickQuestions": 0, "name": ""}], "name": "Luis's copy of Integration by partial fractions", "variable_groups": [], "type": "question", "functions": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "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 ", "parts": [{"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 ", "steps": [{"prompt": "\nFirst 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 $B$.
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", "partialCredit": 0, "strings": ["."]}, "expectedvariablenames": [], "marks": 3, "showCorrectAnswer": true, "checkingaccuracy": 0.001, "vsetrange": [11, 12], "showpreview": true, "answer": "({d-a*c}/{b-a})*ln(x+{a})+({d-b*c}/{a-b})*ln(x+{b})+C", "vsetrangepoints": 5}], "stepsPenalty": 1, "type": "gapfill", "showCorrectAnswer": true, "marks": 0}], "ungrouped_variables": ["a", "c", "b", "d", "s3", "s2", "s1", "b1", "d1"], "metadata": {"notes": "\n \t\t5/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tAdded decimal point as forbidden string.
\n \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\tImproved display of Advice.
\n \t\tAdded information about Show steps, also introduced penalty of 1 mark.
\n \t\tAdded !noLeadingMinus to ruleset std for display purposes.
\n \t\t", "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.
", "licence": "Creative Commons Attribution 4.0 International"}, "tags": ["2 distinct linear factors", "Calculus", "MAS1601", "Steps", "checked2015", "completing the square", "constant of integration", "factorising a quadratic", "indefinite integration", "integration", "logarithms", "partial fractions", "two distinct linear factors"], "showQuestionGroupNames": false, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "variables": {"c": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "c", "definition": "random(2..9)"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "b", "definition": "if(b1=a,b1+s3,b1)"}, "b1": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "b1", "definition": "s2*random(1..9)"}, "d": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "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))"}, "s3": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "s3", "definition": "random(1,-1)"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "s2", "definition": "random(1,-1)"}, "d1": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "d1", "definition": "s3*random(1..9)"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "a", "definition": "s1*random(1..9)"}, "s1": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "s1", "definition": "random(1,-1)"}}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}