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Factorise $x^2+bx+c$ into 2 distinct linear factors and then find $\\displaystyle \\int \\frac{a}{x^2+bx+c }\\;dx$ using partial fractions or otherwise.

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Find the following integral.

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

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Input all numbers as fractions or integers and not decimals.

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Input the constant of integration as $C$.

", "advice": "\n

First we factorise $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+{a})*(x+{b})}$.

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You can do this by spotting the factors or by completing the square.

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Next we use partial fractions to find $A$ and $B$ such that:
\\[\\simplify[std]{{c}/((x +{a})*(x+{b})) = A/(x+{a})+B/(x+{b})}\\]
Multiplying both sides of the equation by $\\displaystyle \\simplify[std]{1/((x +{a})*(x+{b}))}$ we obtain:

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$\\simplify[std]{A*(x+{b})+B*(x+{a}) = {c}}\\Rightarrow \\simplify[std]{(A+B)x+{b}A+{a}B={c}}$

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Identifying coefficients:

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Constant term: $\\simplify[std]{{b}*A+{a}*B = {c}}$

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Coefficent $x$: $ \\simplify[std]{A + B = 0}$ which gives $A = -B$.

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Hence we obtain $\\displaystyle \\simplify[std]{A = {c}/{b-a}}$ and $\\displaystyle \\simplify[std]{B={-c}/{b-a}}$

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Which gives: \\[\\simplify[std]{{c}/((x +{a})*(x+{b})) = ({c}/{b-a})*(1/(x+{a}) -1/(x+{b}))}\\]

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So \\[\\begin{eqnarray*} I &=& \\simplify[std]{Int({c}/(x^2+{a+b}*x+{a*b}),x )}\\\\ &=& \\simplify[std]{Int({c}/((x +{a})*(x+{b})),x )}\\\\ &=& \\simplify[std]{({c}/{b-a})*(Int(1/(x+{a}),x) -Int(1/(x+{b}),x))}\\\\ &=& \\simplify[std]{({c}/{b-a})*(ln(x+{a})-ln(x+{b}))+C} \\end{eqnarray*}\\]

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

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Input all numbers as fractions or integers and not decimals.

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Input the constant of integration as $C$.

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Click on Show steps for help if you need it. You will lose 1 mark if you do so.

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First of all factorise the denominator.

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You have to find $a$ and $b$ such that $\\simplify[std]{x^2+{a+b}*x+{a*b}=(x+a)*(x+b)}$.

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Then use partial fractions to write:
\\[\\simplify[std]{{c}/((x +a)*(x+b)) = A/(x+a)+B/(x+b)}\\]

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for suitable integers or fractions $A$ and $B$.

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Input all numbers as fractions or integers and not decimals.

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