Do not input numbers as decimals, only as integers without the decimal point, or fractions
", "strings": ["."], "partialCredit": 0, "showStrings": false}, "checkingAccuracy": 0.001, "customMarkingAlgorithm": "", "variableReplacements": [], "checkVariableNames": false, "failureRate": 1, "vsetRange": [0, 1], "scripts": {}, "type": "jme", "showCorrectAnswer": true, "answerSimplification": "std", "variableReplacementStrategy": "originalfirst", "marks": 1, "extendBaseMarkingAlgorithm": true}], "showFeedbackIcon": true, "customMarkingAlgorithm": "", "variableReplacements": [], "scripts": {}, "prompt": "$ I = \\ displaystyle \\ int \\ simplify [std] {({b} * x + {c}) / (({a} * x + {d}) ^ {n})} dx $
\nSe le da que \\ [I = \\ simplify [std] {g (x) * ({a} x + {d}) ^ {1-n}} + C \\] para un polinomio $ g (x) $.
\nTienes que encontrar $ g (x) $.
\n$ g (x) = \\; $
Recuerda ingresar todos los números como números enteros o fracciones.
", "type": "gapfill", "sortAnswers": false, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "marks": 0, "extendBaseMarkingAlgorithm": true}], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "statement": "Encuentra la siguiente integral indefinida.
\nIngrese todos los números como números enteros o fracciones, no como decimales.
\n", "functions": {}, "variable_groups": [], "extensions": [], "metadata": {"description": "
$\\displaystyle \\int \\frac{bx+c}{(ax+d)^n} dx=g(x)(ax+d)^{1-n}+C$ for a polynomial $g(x)$. Find $g(x)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "advice": "\nLet $y = \\simplify[std]{{a}*x+{d}}$.
\nThen $x=\\frac{1}{\\var{a}}\\simplify[std]{(y-{d})}$ and so we have the numerator $\\simplify[std]{{b}*x+{c}}$ becomes in terms of $y$:
\n$\\simplify[std]{{b}*x+{c} = {b}*1/{a}*(y-{d})+{c}= {m}y+{r}}$ and so
\n\\[\\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}}\\]