A question on partial fractions

Question

I have the given function which I must convert to partial fractions:

\begin{equation} \frac{x^2}{(x^2+1)(x^2+9)} \end{equation}

and I thought that I should prepare this as:

\begin{equation} \frac{A}{(x+i)}+\frac{Bx}{x^2+1}+\frac{Cx}{x^2+9}+\frac{D}{x+3i} \end{equation}

But this is clearly wrong, because Wolfram says that it must be in the form:

\begin{equation} \frac{A}{x+i}+\frac{B}{x+3i}+\frac{C}{x+i} \end{equation}

Is there a trick on how to identify the correct partial fractions, when the denominator clearly has a factorized complex 4th order polynomial as such?

Thanks

Answer 1

I'm not sure why you are attempting to use complex numbers here. Simply let $u = x^2$ and this becomes:

$$\frac{u}{(u + 1)(u + 9)}$$

For the decomposition, use: $$\frac{A}{u + 1} + \frac{B}{u + 9} = \frac{u}{(u + 1)(u + 9)}$$

This yields:

$$ Au + 9A + Bu + B = u$$

Separating out the $u$ and non-$u$ portions gives us:

$$ Au + Bu = u $$

and

$$ 9A + B = 0 $$

We can solve for $A$ in the first one as: $$ A = 1 - B $$

Then substitute it in:

$$ 9(1 - B) + B = 0 \\ 9 - 9B + B = 0 \\ 9 - 8B = 0 \\ 8B = 9 \\ B = \frac{9}{8}$$

Which means $A = -\frac{1}{8}$.

So your integration becomes:

$$\frac{-\frac{1}{8}}{x^2 + 1} + \frac{\frac{9}{8}}{x^2 + 9}$$

You can then use $arctan$ for the integration.

Answer 2

HINT

Here is another way to approach it for the sake of curiosity \begin{align*} \frac{x^{2}}{(x^{2} + 1)(x^{2} + 9)} & = \frac{(x^{2} + 1) - 1}{(x^{2} + 1)(x^{2} + 9)}\\\\ & = \frac{1}{x^{2} + 9} - \frac{1}{(x^{2} + 1)(x^{2} + 9)} \end{align*}

The last expression, by its turn, can be rewritten as \begin{align*} \frac{1}{(x^{2} + 1)(x^{2} + 9)} & = \frac{1}{8}\times\frac{8}{(x^{2} + 1)(x^{2} + 9)}\\\\ & = \frac{1}{8}\times\frac{(x^{2} + 9) - (x^{2} + 1)}{(x^{2} + 1)(x^{2} + 9)}\\\\ & = \frac{1}{8}\times\left[\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 9}\right] \end{align*}

Hence the original expression can be expressed as \begin{align*} \frac{x^{2}}{(x^{2} + 1)(x^{2} + 9)} = -\frac{1}{8(x^{2} + 1)} + \frac{9}{8(x^{2} + 9)} \end{align*}

Can you take it from here?

Answer 3

The 'right' partial fraction decomposition is rather \begin{eqnarray*} \phi(x) &=& \frac{x^2}{(x^2+1)(x^2+9)} \\ &=& \frac{x^2}{(x-i)(x+i)(x-3i)(x+3i)}\\ &=& \frac{A}{x-i}+ \frac{B}{x+i}+ \frac{C}{x-3i}+ \frac{D}{x+3i} \end{eqnarray*} It holds $$ A = [\phi(x)(x-i)]_{x=+i} = \frac{i^2}{(i+i)(i^2+9)} = \frac{1}{8} \cdot \frac{i}{2} $$ and its conjugate $$ B = [\phi(x)(x+i)]_{x=-i} = \frac{(-i)^2}{(-i-i)((-i)^2+9)} = \frac{1}{8} \cdot \frac{-i}{2} $$ The rest is similar.