Input all numbers as fractions or integers and not decimals.
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t$I=\\;$[[0]]
\n\t\t\tInput all numbers as fractions or integers and not decimals.
\n\t\t\tInput the constant of integration as $C$.
\n\t\t\tClick on Show steps for help if you need it. You will lose 1 mark if you do so.
\n\t\t\t", "steps": [{"type": "information", "prompt": "\n\t\t\t\t\tFirst of all factorise the denominator.
\n\t\t\t\t\tYou have to find $a$ and $b$ such that $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+a)*(x+b)}$.
\n\t\t\t\t\tThen use partial fractions to write:
\\[\\simplify[std]{{c}/((x +a)*(x+b)) = A/(x+a)+B/(x+b)}\\]
for suitable integers or fractions $A$ and $B$.
\n\t\t\t\t\t", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\n\tFind the following integral.
\n\t\\[I = \\simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}\\]
\n\tInput all numbers as fractions or integers and not decimals.
\n\tInput the constant of integration as $C$.
\n\t", "tags": ["2 distinct linear factors", "Calculus", "MAS1601", "Steps", "checked2015", "completing the square", "constant of integration", "factorising a quadratic", "factorizing a quadratic", "indefinite integration", "integration", "partial fractions", "two distinct linear factors"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "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", "licence": "Creative Commons Attribution 4.0 International", "description": "\n\t\tFactorise $x^2+bx+c$ into 2 distinct linear factors and then find $\\displaystyle \\int \\frac{a}{x^2+bx+c }\\;dx$ using partial fractions or otherwise.
\n\t\t\n\t\t"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t
First we factorise $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+{a})*(x+{b})}$.
\n\tYou can do this by spotting the factors or by completing the square.
\n\tNext we use partial fractions to find $A$ and $B$ such that:
\\[\\simplify[std]{{c}/((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:
$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}}\\Rightarrow \\simplify[std]{(A+B)x+{b}A+{a}B={c}}$
\n\tIdentifying coefficients:
\n\tConstant term: $\\simplify[std]{{b}*A+{a}*B = {c}}$
\n\tCoefficent $x$: $ \\simplify[std]{A + B = 0}$ which gives $A = -B$.
\n\tHence we obtain $\\displaystyle \\simplify[std]{A = {c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={-c}/{b-a}}$
\n\tWhich gives: \\[\\simplify[std]{{c}/((x +{a})*(x+{b})) = ({c}/{b-a})*(1/(x+{a}) -1/(x+{b}))}\\]
\n\tSo \\[\\begin{eqnarray*} I &=& \\simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}\\\\ &=& \\simplify[std]{Int({c}/((x +{a})*(x+{b})),x )}\\\\ &=& \\simplify[std]{({c}/{b-a})*(Int(1/(x+{a}),x) -Int(1/(x+{b}),x))}\\\\ &=& \\simplify[std]{({c}/{b-a})*(ln(x+{a})-ln(x+{b}))+C} \\end{eqnarray*}\\]
\n\t", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}