Simon's copy of Integration by partial fractions

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Question 1

Find the following integral.

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

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

$I=\;$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{-\ln \left ( x - 2 \right ) + 6 \ln \left ( x - 4 \right ) + C}$

Input all numbers as fractions or integers and not decimals.

Input the constant of integration as $C$ .

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

Use partial fractions in order to write:
$\simplify[std]{({c}*x+{d})/((x +{a})*(x+{b}))}\; = \simplify[std]{A/(x+{a})+B/(x+{b})}$

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

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Advice

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

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

Identifying coefficients:

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

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

On solving these equations we obtain $\displaystyle \simplify[std]{A = {d-a*c}/{b-a}}$ and $\displaystyle \simplify[std]{B={d-b*c}/{a-b}}$

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

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

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Simon's copy of Integration by partial fractions

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