Question

How do you express $\dfrac{1}{{{x}^{4}}+1}$ in partial fractions?

Answer 1

Hint: In this question, the denominator is a polynomial that is ${{x}^{4}}+1$ and has not a real root, therefore it cannot be real factorized then it cannot be expanded in real partial fraction. Therefore, to solve this we have to factor it into quadratic factors and then solve it in partial fraction.

Complete step by step solution:
Transforming a rational polynomial into a sum of simpler rational polynomials due to factorization of denominator is called partial fraction.
We have our expression as:
$\Rightarrow \dfrac{1}{{{x}^{4}}+1}$
Since, the denominator is a polynomial that is ${{x}^{4}}+1$ has no linear factors with Real coefficients because ${{x}^{4}}+1\ge 1>0$ for all Real values of $x$.
But, it is possible to factor it into quadratic factors:
Therefore, we get:
$\Rightarrow {{x}^{4}}+1=\left( {{x}^{2}}-\sqrt{2}+1 \right)\left( {{x}^{2}}+\sqrt{2}x+1 \right)$
With this, we can do partial fraction decomposition in the form:
$\Rightarrow \dfrac{1}{{{x}^{4}}+1}=\dfrac{Ax+B}{{{x}^{2}}-\sqrt{2}+1}+\dfrac{Cx+D}{{{x}^{2}}+\sqrt{2}x+1}$
On simplifying left-hand side and right-hand side, we get:
$\Rightarrow 1=\dfrac{\left( Ax+B \right)\left( {{x}^{2}}+\sqrt{2}x+1 \right)+\left( Cx+D \right)\left( {{x}^{2}}-\sqrt{2}x+1 \right)}{{{x}^{4}}+1}$
On further expanding and simplifying, we get:
$\Rightarrow \dfrac{\left( A+C \right){{x}^{3}}+\left( \sqrt{2}A+B-\sqrt{2}C+D \right){{x}^{3}}+\left( A+\sqrt{2}B+C-\sqrt{2}D \right)x+\left( B+D \right)}{{{x}^{4}}+1}$
Now, we will equate coefficients, so we get:
$\begin{align}
  & A+C=0\to \left( 1 \right) \\
 & \sqrt{2}A+B-\sqrt{2}C+D=0\to \left( 2 \right) \\
 & A+\sqrt{2}B+C-\sqrt{2}D=0\to \left( 3 \right) \\
 & B+D=1\to \left( 4 \right) \\
\end{align}$
Now, we will subtract $\left( 1 \right)$ from $\left( 3 \right)$:
So, we get:
$A+\sqrt{2}B+C-\sqrt{2}D-A-C=0$
Now we will divide by $\sqrt{2}$to clear the equation.
Therefore, on simplifying, we get:
$A+B+C-D-A-C=0$
$B-D=0$
Therefore, we get:
$B=D$
Now, we will substitute this with the $\left( 4 \right)$ equation.
Therefore, we get:
$B+B=1$
$2B=1$
On simplifying, we get:
$\Rightarrow B=D=\dfrac{1}{2}$
Now, substituting $B+D=1$ in the $\left( 2 \right)$ equation, we get:
$\Rightarrow \sqrt{2}A-\sqrt{2}C+1=0$
Now combining this with $\left( 1 \right)$ equation, we get:
$\Rightarrow 2\sqrt{2}A=-1$
Therefore,
$\Rightarrow A=-\dfrac{\sqrt{2}}{4}$ and $C=\dfrac{\sqrt{2}}{4}$
With this, now substituting our solution in actual practical fraction, we get our final answer that is:
$\Rightarrow \dfrac{1}{{{x}^{4}}+1}=\dfrac{-\sqrt{2}x+2}{{{x}^{2}}-\sqrt{2}+1}+\dfrac{\sqrt{2}x+2}{{{x}^{2}}+\sqrt{2}+1}$

Note: It is to be remembered that the partial fraction skeleton is different for different values of the polynomial expressions. The main application of using partial fractions is in integration when the integration has to be performed on complex polynomial fractions which cannot be directly integrated.