Step by step solving for integration by substitution
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\nDon't forget the constant of integration ($C$).
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\n\nUse $u=\\simplify[std]{{a[0]}x^2+{b[0]}}$ as your substitution.
\n$\\frac{du}{dx}=$ [[1]]
\n$dx=$ [[2]]
\nSubstituting back into the original equation for $dx$ and pulling out constants gives
\n$I=$[[3]]$\\simplify[std]{Int(u^{m[0]},u)}$
\nThe next step is to integrate.
\n$\\simplify{Int(u^{m[0]},u)}=$ [[4]]
\nPutting all of these results together, we get the final answer of:
\n[[0]]
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", "showStrings": false, "strings": ["."], "partialCredit": 0}, "checkingAccuracy": 0.001, "scripts": {}, "answer": "(({a[0]}*(x^2)+{b[0]})^({m[0]}+1))/(2*{a[0]}*({m[0]}+1))+C", "valuegenerators": [{"value": "", "name": "c"}, {"value": "", "name": "x"}], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": "2", "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}, {"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": false, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "2{a[0]}x", "valuegenerators": [{"value": "", "name": "x"}], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": 1, "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}, {"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": false, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "du/(2{a[0]}x)", "valuegenerators": [{"value": "", "name": "du"}, {"value": "", "name": "x"}], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": "1", "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}, {"customName": "", "customMarkingAlgorithm": "", "correctAnswerStyle": "plain", "scripts": {}, "maxValue": "1/(2{a[0]})", "allowFractions": true, "mustBeReducedPC": 0, "extendBaseMarkingAlgorithm": true, "showFractionHint": true, "showCorrectAnswer": false, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "showFeedbackIcon": true, "notationStyles": ["plain", "en", "si-en"], "marks": 1, "unitTests": [], "minValue": "1/(2{a[0]})", "useCustomName": false, "correctAnswerFraction": true, "variableReplacements": [], "mustBeReduced": false}, {"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": false, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "u^({m[0]}+1)/({m[0]}+1)", "valuegenerators": [{"value": "", "name": "u"}], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": "1", "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}], "variableReplacements": []}], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "advice": "This problem is best solved by using substitution.
Note that if we let $u=\\simplify[std]{{a[0]} * (x ^ 2) + {b[0]}}$ then $du=\\simplify[std]{(2*{a[0]} * x)*dx }$
Hence we can replace $xdx$ by $\\frac{1}{2*\\var{a[0]}}du$.
Hence the integral becomes:
\n\\[\\begin{eqnarray*} I&=&\\simplify[std]{Int((1/(2{a[0]}))u^{m[0]},u)}\\\\ &=&\\simplify[all]{(1/(2{a[0]}))u^{m[0]+1}/{m[0]+1}+C}\\\\ &=& \\simplify[all]{({a[0]} * (x ^ 2) + {b[0]})^{m[0]+1}/(2{a[0]}*({m[0]}+1))+C} \\end{eqnarray*}\\]
\nA Useful Result
This example can be generalised.
Suppose \\[I = \\int\\; f'(x)g(f(x))\\;dx\\]
The using the substitution $u=f(x)$ we find that $du=f'(x)\\;dx$ and so using the same method as above:
\\[I = \\int g(u)\\;du \\]
And if we can find this simpler integral in terms of $u$ we can replace $u$ by $f(x)$ and get the result in terms of $x$.