Do not input numbers as decimals, only as integers without the decimal point, or fractions
", "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$I=\\displaystyle \\int \\simplify[std]{({b}*x+{c})/(({a}*x+{d})^{n})} dx$
\n\t\t\tYou are given that \\[I=\\simplify[std]{g(x)*({a}x+{d})^{1-n}}+C\\] for a polynomial $g(x)$. You have to find $g(x)$.
\n\t\t\t$g(x)=\\;$[[0]]
\n\t\t\tRemember to input all numbers as integers or fractions.
\n\t\t\tClick on Show steps to get help if you need it. You will lose 1 mark by doing so.
\n\t\t\t", "steps": [{"type": "information", "prompt": "One way to do this is by substitution, for example $y = \\simplify[std]{{a}*x+{d}}$.
", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "\n\tFind the following indefinite integral.
\n\tInput all numbers as integers or fractions, not as decimals.
\n\tInput the constant of integration as $C$.
\n\t", "tags": ["Calculus", "MAS1601", "Steps", "checked2015", "constant of integration", "indefinite integration", "integration", "integration by substitution", "substitution"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t2/08/2012:
\n\t\tAdded tags.
\n\t\tAdded description.
\n\t\tAdded a Step and message about Show steps included - losing 1 mark if used as it gives the formula for finding the integral. Increased marks to 3 for the question, so that can cope with losing a mark for using Show steps.
\n\t\tChecked calculation. OK.
\n\t\tImproved display in Advice.
\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Find the polynomial $g(x)$ such that $\\displaystyle \\int \\frac{ax+b}{(cx+d)^{n}} dx=\\frac{g(x)}{(cx+d)^{n-1}}+C$.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\tLet $y = \\simplify[std]{{a}*x+{d}}$.
\n\tThen $\\displaystyle x=\\frac{1}{\\var{a}}\\simplify[std]{(y-{d})}$ and so we have the numerator $\\simplify[std]{{b}*x+{c}}$ becomes in terms of $y$:
\n\t$\\displaystyle \\simplify[std]{{b}*x+{c} = {b}*1/{a}*(y-{d})+{c}= {m}y+{r}}$ and so
\n\t\\[\\displaystyle \\simplify[std]{({b}*x+{c})/(({a}*x+{d})^{n})} = \\simplify[std]{({m}*y+{r})/(y^{n})={m}/y^{n-1}+{r}/y^{n}}\\]
\n\tNow,
\\[\\int \\simplify[std]{({b}x+{c})/({a}*x+{d})^{n}} dx = \\int \\left(\\simplify[std]{{m}/y^{n-1}+{r}/y^{n}} \\right)\\frac{dx}{dy} dy \\]
Since $\\displaystyle x = \\simplify[std]{(y-{d})/{a}}$ then $\\displaystyle \\frac{dx}{dy} = \\frac{1}{\\var{a}}$.
\n\tWe can now calculate the desired integral:
\n\t\\[ \\begin{eqnarray*} \\int \\left(\\simplify[std]{{m}/y^{n-1}+{r}/y^{n}}\\right) \\frac{dx}{dy} dy &=&\\frac{1}{\\var{a}}\\left(\\int \\simplify[std]{{m}/y^{n-1}}\\;dy+\\int \\simplify[std]{{r}/y^{n}}\\;dy \\right)\\\\ &=&\\frac{1}{\\var{a}}\\left(\\simplify[std]{{-m}/({n-2}*y^{n-2})+ {-r}/({n-1}*y^{n-1})}\\right) + C \\\\ &=& \\simplify[std]{(-{m})/({a*(n-2)}*({a}*x+{d})^{n-2})+(-{r})/({a*(n-1)}*({a}*x+{d})^{n-1}) + C}\\\\ &=&\\simplify[std]{1/({a}x+{d})^{n-1}*(({-m}/{a*(n-2)})*({a}x+{d})-{r}/({a*(n-1)}))}\\\\ &=&\\simplify[std]{1/({a}x+{d})^{n-1}*(({-m}/{(n-2)})*x+{-m*d*(n-1)-r*(n-2)}/{(n-2)*(n-1)*a})} \\end{eqnarray*} \\]
Hence \\[g(x)=\\simplify[std]{({-m}/{(n-2)})*x+{-m*d*(n-1)-r*(n-2)}/{(n-2)*(n-1)*a}}\\]