Question

How do I find the partial decomposition of $ \dfrac{{{x^4} + 1}}{{{x^5} + 4{x^3}}}? $

Answer 1

Hint: Partial fraction decomposition or partial fraction expansion of a rational fraction is an operation in which the denominator is expressed as a sum of a polynomial and one or several fractions with a simpler denominator.

Complete step by step solution:
Given $ \dfrac{{{x^4} + 1}}{{{x^5} + 4{x^3}}} $ .
The motto of this question is, how we can partially decompose the given equation.
As said in hint, we must first factor the denominator as much as possible,
So, let us take $ {x^3} $ in denominator as common and write as follows $ : $
 $ \Rightarrow \dfrac{{{x^4} + 1}}{{{x^3}({x^2} + 4)}} $
Now, let us choose the factors to be written as $ : $
 $ \Rightarrow \dfrac{{{x^4} + 1}}{{{x^3}({x^2} + 4)}} = \dfrac{A}{x} + \dfrac{B}{{{x^2}}} + \dfrac{C}{{{x^3}}} + \dfrac{{Dx + E}}{{{x^2} + 4}} - - - (1) $
Here, $ {x^3} $ is written as successive powers of $ x $ , starting at $ x $ followed by $ {x^2} $ and then ending with $ {x^3} $ . Also, there was a quadratic term $ {x^2} + 4 $ which could not be factored. Hence, we used $ Dx + E $ in the numerator.
Next, let us multiply \[{x^3}({x^2} + 4)\] on both sides of the equation,
\[ \Rightarrow \dfrac{{{x^3}({x^2} + 4)({x^4} + 1)}}{{{x^3}({x^2} + 4)}} = \dfrac{{{x^3}({x^2} + 4)A}}{x} + \dfrac{{{x^3}({x^2} + 4)B}}{{{x^2}}} + \dfrac{{{x^3}({x^2} + 4)C}}{{{x^3}}} + \dfrac{{{x^3}({x^2} + 4)(Dx + E)}}{{{x^2} + 4}}\]
After cancelling all the common terms in the above equation, we get:
 $ {x^4} + 1 = A{x^2}({x^2} + 4) + Bx({x^2} + 4) + C({x^2} + 4) + {x^3}(Dx + E) $
On simplifying further,
\[{x^4} + 1 = A{x^4} + 4A{x^2} + B{x^3} + 4Bx + C{x^2} + 4C + D{x^4} + E{x^3}\]
By using the technique of grouping, we will now solve for each constant.
The first step is to rearrange every term in successive powers of $ x $ .
\[{x^4} + 1 = (A + D){x^4} + (B + E){x^3} + (4A + C){x^2} + 4Bx + 4C\]
Next, we create a system of equation using the coefficients of $ x $ on the left side that correspond to the coefficients of x on the right side.
That is,
Coefficient of $ {x^4} \to A + D = 1 $
Coefficient of $ {x^3} \to B + E = 0 $
Coefficient of $ {x^2} \to 4A + C = 0 $
Coefficient of $ {x^1} = x \to 4B = 0 $
Coefficient of $ {x^0} \to 4C = 1 $
From the above system of equations, we can see that $ B = 0 $ and $ C = \dfrac{1}{4} $ .
From these it follows that since $ B + E = 0 $ and $ B = 0 $ , $ E = 0 $ .
And since $ 4A + C = 0 $ and $ C = \dfrac{1}{4} $ ,
 $ 4A + \dfrac{1}{4} = 0 $
 $ \Rightarrow 4A = - \dfrac{1}{4} $
On further simplifying,
 $ A = - \dfrac{1}{{16}} $
Finally, substituting $ A $ in $ A + D = 1 $ , we get
 $ - \dfrac{1}{{16}} + D = 1 $
 $ \Rightarrow D = 1 + \dfrac{1}{{16}} $
Thus, $ D = \dfrac{{17}}{{16}} $
Now, substituting these values in equation $ (1) $ , we get
 $ \Rightarrow \dfrac{{{x^4} + 1}}{{{x^3}({x^2} + 4)}} = - \dfrac{1}{{16x}} + \dfrac{0}{{{x^2}}} + \dfrac{1}{{4{x^3}}} + \dfrac{{\dfrac{{17}}{{16}}x + 0}}{{{x^2} + 4}} $
 $ \Rightarrow \dfrac{{{x^4} + 1}}{{{x^3}({x^2} + 4)}} = - \dfrac{1}{{16x}} + \dfrac{1}{{4{x^3}}} + \dfrac{{17x}}{{16({x^2} + 4)}} $
Rearranging the above equation, we get
 $ \Rightarrow \dfrac{{{x^4} + 1}}{{{x^3}({x^2} + 4)}} = \dfrac{1}{{4{x^3}}} + \dfrac{{17x}}{{16({x^2} + 4)}} - \dfrac{1}{{16x}} $
This is the required solution.
So, the correct answer is “ $ \dfrac{{{x^4} + 1}}{{{x^3}({x^2} + 4)}} = \dfrac{1}{{4{x^3}}} + \dfrac{{17x}}{{16({x^2} + 4)}} - \dfrac{1}{{16x}} $ ”.

Note: Expanding with partial fractions is all about choosing the correct factors. If correct factors are chosen and grouping technique is known, then solving such questions is easy.