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# How do you express $\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}$ in partial fractions?

Last updated date: 13th Jun 2024
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Hint: The denominator of the expression is the product of a linear expression i.e. $\left( {x - 1} \right)$ and a quadratic expression i.e. $\left( {{x^2} + 4} \right)$ and its degree is also greater than that of numerator. Thus the partial fraction will be in the form of $\dfrac{A}{{\left( {x - 1} \right)}} + \dfrac{{Bx + C}}{{\left( {{x^2} + 4} \right)}}$. Compare this with the initial expression given in the question to find the values of $A,B$ and $C$.

According to the question, we have to show how to resolve the given algebraic expression in partial fractions.
The algebraic expression is $\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}$.
If we observe the denominator of the expression, it is the product of a linear and a quadratic expression and its degree is also greater than the degree of the numerator. We know that in such cases, the partial fractions will be expressed as:
$\Rightarrow \dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \dfrac{A}{{\left( {x - 1} \right)}} + \dfrac{{Bx + C}}{{\left( {{x^2} + 4} \right)}}{\text{ }}.....{\text{(1)}}$
From here, we just have to calculate the values of $A,B$ and $C$. We will do so by comparing the coefficients on both sides of the equation.
On further simplification, equation (1) will give us:
$\Rightarrow \dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{A\left( {{x^2} + 4} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}$
Denominators on both sides will cancel out. We’ll get:
$\Rightarrow x = A\left( {{x^2} + 4} \right) + \left( {Bx + C} \right)\left( {x - 1} \right) \\ \Rightarrow x = A{x^2} + 4A + B{x^2} - Bx + Cx - C \\$
Arranging the terms, this can be further written as:
$\Rightarrow x = \left( {A + B} \right){x^2} + \left( {C - B} \right)x + 4A - C$
Comparing coefficients of ${x^2}$ on both sides, we’ll get:
$\Rightarrow A + B = 0 \\ \Rightarrow A = - B{\text{ }}.....{\text{(2)}} \\$
Similarly comparing coefficients of $x$ on both sides, we’ll get:
$\Rightarrow C - B = 1 \\ \Rightarrow C = B + 1 \\$
Putting $B = - A$ here from equation (2), we have:
$\Rightarrow C = 1 - A{\text{ }}.....{\text{(3)}}$
And finally comparing constant terms on both sides, we’ll get:
$\Rightarrow 4A - C = 0 \\ \Rightarrow C = 4A \\$
Putting $C = 1 - A$ in this from equation (3), we’ll get:
$\Rightarrow 1 - A = 4A \\ \Rightarrow 5A = 1 \\ \Rightarrow A = \dfrac{1}{5} \\$
We have already calculated $B = - A$. SO we have:
$\Rightarrow B = - \dfrac{1}{5}$
Using $C = 1 - A$ again from equation (3) and putting $A = \dfrac{1}{5}$, we have:
$\Rightarrow C = 1 - \dfrac{1}{5} \\ \Rightarrow C = \dfrac{4}{5} \\$
Thus we finally have $A = \dfrac{1}{5},{\text{ }}B = - \dfrac{1}{5}$ and $C = \dfrac{4}{5}$. Putting these values in equation (1), we’ll get:
$\Rightarrow \dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \dfrac{1}{{5\left( {x - 1} \right)}} + \dfrac{{ - \dfrac{1}{5}x + \dfrac{4}{5}}}{{\left( {{x^2} + 4} \right)}} \\ \Rightarrow \dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \dfrac{1}{{5\left( {x - 1} \right)}} + \dfrac{{4 - x}}{{5\left( {{x^2} + 4} \right)}} \\$
This is the required partial fraction of the expression.

Note: Some other important algebraic expressions along with their partial fractions are given below:
(1) $\dfrac{k}{{\left( {x - a} \right)\left( {x - b} \right)\left( {x - c} \right)...}} = \dfrac{A}{{\left( {x - a} \right)}} + \dfrac{B}{{\left( {x - b} \right)}} + \dfrac{C}{{\left( {x - c} \right)}} + ...$
(2) $\dfrac{k}{{\left( {x - a} \right){{\left( {x - b} \right)}^2}}} = \dfrac{A}{{\left( {x - a} \right)}} + \dfrac{B}{{\left( {x - b} \right)}} + \dfrac{C}{{{{\left( {x - b} \right)}^2}}}$
(3) $\dfrac{{p{x^2} + qx + c}}{{\left( {x - a} \right){{\left( {x - b} \right)}^2}}} = \dfrac{A}{{\left( {x - a} \right)}} + \dfrac{B}{{\left( {x - b} \right)}} + \dfrac{C}{{{{\left( {x - b} \right)}^2}}}$
The procedure to solve is the same for all the partial fractions. We have to compare the coefficients on both sides.
While solving partial fraction one more thing to keep in mind is that the degree of the numerator must be less than that of the denominator. If this is not the case then we have to reduce the expression such that the above condition satisfies. For example,
$\Rightarrow \dfrac{{{x^2}}}{{{{\left( {x + 1} \right)}^2}}} = 1 - \dfrac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}$
Here we can see that initially the degree of both numerator and denominator was 2. But we reduced it to another form to meet our condition. Now $\dfrac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}$ can be easily converted into partial fractions.