\\[I=\\simplify[std]{Int(({2 * a} * x + {b}) / ({a} * x ^ 2 + {b} * x + {c}),x)}\\]

\n$I=\\;$[[0]]

\nInput the constant of integration as $C$.

\nInput all numbers as integers or fractions not as decimals.

\nClick on Show steps if you need help. You will lose 1 mark if you do so.

", "type": "gapfill", "sortAnswers": false, "scripts": {}, "marks": 0, "steps": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "customMarkingAlgorithm": "", "showFeedbackIcon": true, "marks": 0, "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "unitTests": [], "prompt": "Try the substitution $u=\\simplify[std]{{a} * (x ^ 2) + ({b} * x) + {c}}$

", "type": "information"}], "unitTests": [], "stepsPenalty": 1}], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "variable_groups": [], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Find $\\displaystyle \\int \\frac{2ax + b}{ax ^ 2 + bx + c}\\;dx$

"}, "name": "Indefinite integral by substitution 3 (custom feedback)", "advice": "\n\tThis exercise is best solved by using substitution.

\n\tNote that the numerator $\\simplify[std]{{2 * a} * x + {b}}$ of \\[\\simplify[std]{({2 * a} * x + {b}) / ({a} * x ^ 2 + {b} * x + {c})}\\] is the derivative of the denominator $\\simplify[std]{{a} * x ^ 2 + {b} * x + {c}}$

\n\tSo if you use as your substitution $u=\\simplify[std]{{a} * (x ^ 2) + ({b} * x) + {c}}$ you then have $\\simplify[std]{ du = ({2 * a} * x + {b}) * dx}$

\n\tHence we can replace $\\simplify[std]{ ({2 * a} * x + {b}) * dx}$ by $du$

\n\tHence the integral becomes:

\n\t\\[\\begin{eqnarray*} I&=&\\int\\;\\frac{du}{u}\\\\ &=&\\ln(|u|)+C\\\\ &=& \\simplify[std]{ln(abs({a} * (x ^ 2) + ({b} * x) + {c}))+C} \\end{eqnarray*}\\]

**A Useful Result**

This example can be generalised.

Suppose \\[I = \\int\\; \\frac{f'(x)}{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 \\frac{du}{u} = \\ln(|u|)+ C = \\ln(|f(x)|)+C\\]

Find the following integral.

\n\tInput the constant of integration as $C$.

\n\tInput all numbers as integers or fractions.

\n\t\n\t \n\t \n\t", "type": "question", "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}]}]}], "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}]}