Question

How do you express $\dfrac{{{x^4} + 6}}{{{x^5} + 7{x^3}}}$ in partial fractions?

Answer 1

Hint : Partial fractions is the method of decomposing the given fraction into its initial polynomial fraction. For partial fractions first we factorize the denominator ${x^5} + 7{x^3}$ in this case and then express the given fraction into the sum of the factors each having one factor of the denominator as its example as well as an coefficient after simplifying and comparing it to the original numerator. For an initial hint see this equation and solve it for the coefficients that are $K,L,M.N,O$
$\dfrac{{{x^4} + 6}}{{{x^5} + 7{x^3}}} = \dfrac{K}{x} + \dfrac{L}{{{x^2}}} + \dfrac{M}{{{x^3}}} + \dfrac{{Nx + O}}{{{x^2} + 7}}$

Complete step-by-step answer :
We are given,
$\dfrac{{{x^4} + 6}}{{{x^5} + 7{x^3}}}$
For calculating the partial fractions first we factorize the denominator the denominator can be factorized as follows:
${x^5} + 7{x^3} = {x^3}({x^2} + 7)$ , we should also remember that $x$ and ${x^2}$ are also the factor now we express our partial fraction into its factor and coefficients $K,L,M.N,O$ also remember that since $({x^2} + 7)$ is quadratic its numerator will be taken as a linear instead of constant so we write
$\dfrac{{{x^4} + 6}}{{{x^5} + 7{x^3}}} = \dfrac{K}{x} + \dfrac{L}{{{x^2}}} + \dfrac{M}{{{x^3}}} + \dfrac{{Nx + O}}{{{x^2} + 7}}$
Upon further solving we can write the above equation as
$\dfrac{{{x^4} + 6}}{{{x^5} + 7{x^3}}} = \dfrac{{K{x^2}({x^2} + 7) + Lx({x^2} + 7) + M({x^2} + 7) + {x^3}(Nx + O)}}{{{x^3}({x^2} + 7)}}$
Therefore
${x^4} + 6$ = $K{x^2}({x^2} + 7) + Lx({x^2} + 7) + M({x^2} + 7) + {x^3}(Nx + O)$
Now easy way to solve is to put $x = 0$ we get
$M = \dfrac{6}{7}$
Coefficients of ${x^2}$ putting zero we get $K = \dfrac{{ - 6}}{{49}}$ upon further putting coefficients of ${x^4}$ as $0$ we get
$N = \dfrac{{55}}{{49}}$
Further solving we get $L = 0$ and $O = 0$
Thus we can finally write the given question in partial fractions with the help of the coefficients as follows:
$\dfrac{{{x^4} + 6}}{{{x^5} + 7{x^3}}} = \dfrac{{ - \dfrac{6}{{49}}}}{x} + \dfrac{0}{{{x^2}}} + \dfrac{{\dfrac{6}{7}}}{{{x^3}}} + \dfrac{{\dfrac{{55}}{{49}}x + 0}}{{{x^2} + 7}}$
The above equation is arrived at by putting the various values of coefficients into the equation
$\dfrac{{{x^4} + 6}}{{{x^5} + 7{x^3}}} = \dfrac{K}{x} + \dfrac{L}{{{x^2}}} + \dfrac{M}{{{x^3}}} + \dfrac{{Nx + O}}{{{x^2} + 7}}$

Note : The partial fraction’s denominator if it has any quadratic expression as its factor then we have to take a linear expression as the numerator.