Question

Find the coefficients of partial fraction for the equation $\int {\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)\left( {x - 5} \right)\left( {x - 6} \right)}}}$.

Answer 1

Hint : First of all, split the equation into the desired form of the partial fractions. Then find the unknown by using a suitable method and hence to obtain the required answer. As they have asked to find the coefficients of partial fraction only we need not to solve the whole integral.

Complete step by step solution :
Let $I = \int {\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)\left( {x - 5} \right)\left( {x - 6} \right)}}} dx$
Now, consider the partial fractions of $\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)\left( {x - 5} \right)\left( {x - 6} \right)}}$ i.e.,

$$\Rightarrow \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)\left( {x - 5} \right)\left( {x - 6} \right)}} = \dfrac{A}{{x - 4}} + \dfrac{B}{{x - 5}} + \dfrac{C}{{x - 6}} \\\ \Rightarrow \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)\left( {x - 5} \right)\left( {x - 6} \right)}} = \dfrac{{A\left( {x - 5} \right)\left( {x - 6} \right) + B\left( {x - 4} \right)\left( {x - 6} \right) + C\left( {x - 4} \right)\left( {x - 5} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 6} \right)}} \\\$$

Cancelling the common terms, we have
$\Rightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) = A\left( {x - 5} \right)\left( {x - 6} \right) + B\left( {x - 4} \right)\left( {x - 6} \right) + C\left( {x - 4} \right)\left( {x - 5} \right)$
Put $x = 4$

$$\Rightarrow \left( {4 - 1} \right)\left( {4 - 2} \right)\left( {4 - 3} \right) = A\left( {4 - 5} \right)\left( {4 - 6} \right) + B\left( {4 - 4} \right)\left( {4 - 6} \right) + C\left( {4 - 4} \right)\left( {4 - 5} \right) \\\ \Rightarrow 3 \times 2 \times 1 = A\left( { - 1} \right)\left( { - 2} \right) + 0 + 0 \\\ \Rightarrow 6 = 2A \\\ \therefore A = 3 \\\$$

Put $x = 5$

$$\Rightarrow \left( {5 - 1} \right)\left( {5 - 2} \right)\left( {5 - 3} \right) = A\left( {5 - 5} \right)\left( {5 - 6} \right) + B\left( {5 - 4} \right)\left( {5 - 6} \right) + C\left( {5 - 4} \right)\left( {5 - 5} \right) \\\ \Rightarrow 4 \times 3 \times 2 = 0 + B\left( 1 \right)\left( { - 1} \right) + 0 \\\ \Rightarrow 24 = - B \\\ \therefore B = - 24 \\\$$

Put $x = 6$

$$\Rightarrow \left( {6 - 1} \right)\left( {6 - 2} \right)\left( {6 - 3} \right) = A\left( {6 - 5} \right)\left( {6 - 6} \right) + B\left( {6 - 4} \right)\left( {6 - 6} \right) + C\left( {6 - 4} \right)\left( {6 - 5} \right) \\\ \Rightarrow 5 \times 4 \times 3 = 0 + 0 + C\left( 2 \right)\left( 1 \right) \\\ \Rightarrow 60 = 2C \\\ \therefore C = 30 \\\$$

Thus, the coefficients of partial fraction of $\int {\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 4} \right)\left( {x - 5} \right)\left( {x - 6} \right)}}}$ are $3, - 24,30$.

Note : The partial fraction of the equation of the form $\dfrac{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {x - p} \right)\left( {x - q} \right)\left( {x - r} \right)}}$ is given by $\dfrac{A}{{x - p}} + \dfrac{B}{{x - q}} + \dfrac{C}{{x - r}}$ where $A,B,C$ can be obtained by substituting by $x = p,q,r$ respectively.