Let $y = \\simplify[std]{{a}*x+{d}}$.
\nThen $\\displaystyle x=\\frac{1}{\\var{a}}\\simplify[std]{(y-{d})}$ and so we have the numerator $\\simplify[std]{{b}*x+{c}}$ becomes in terms of $y$:
\n$\\displaystyle \\simplify[std]{{b}*x+{c} = {b}*1/{a}*(y-{d})+{c}= {m}y+{r}}$ and so
\n\\[\\displaystyle \\simplify[std]{({b}*x+{c})/(({a}*x+{d})^{n})} = \\simplify[std]{({m}*y+{r})/(y^{n})={m}/y^{n-1}+{r}/y^{n}}\\]
\nNow,
\\[\\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 $\\displaystyle \\frac{dx}{dy} = \\frac{1}{\\var{a}}$.
\nWe can now calculate the desired integral:
\n\\[ \\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}}\\]
$I=\\displaystyle \\int \\simplify[std]{({b}*x+{c})/(({a}*x+{d})^{n})} dx$
\nYou 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$g(x)=\\;$[[0]]
\nRemember to input all numbers as integers or fractions.
\nClick on Show steps to get help if you need it. You will lose 1 mark by doing so.
\n ", "gaps": [{"notallowed": {"message": "Do not input numbers as decimals, only as integers without the decimal point, or fractions
", "showstrings": false, "strings": ["."], "partialcredit": 0.0}, "checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "({-m}/{n-2})*x-{m*d*(n-1)+r*(n-2)}/{(n-2)*(n-1)*a}", "type": "jme"}], "steps": [{"prompt": "One way to do this is by substitution, for example $y = \\simplify[std]{{a}*x+{d}}$.
", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "extensions": [], "statement": "\nFind the following indefinite integral.
\nInput all numbers as integers or fractions, not as decimals.
\nInput the constant of integration as $C$.
\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..5)", "name": "a"}, "c": {"definition": "m*d+r", "name": "c"}, "b": {"definition": "m*a", "name": "b"}, "d": {"definition": "random(1..5)", "name": "d"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "m": {"definition": "s1*random(1..4)", "name": "m"}, "n": {"definition": "random(3..5)", "name": "n"}, "r": {"definition": "s2*random(1..5)", "name": "r"}}, "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", "description": "Find the polynomial $g(x)$ such that $\\displaystyle \\int \\frac{ax+b}{(cx+d)^{n}} dx=\\frac{g(x)}{(cx+d)^{n-1}}+C$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}]}