How do you use partial fraction decomposition to decompose the fraction to integrate 1/((x^4) +1)?

Jun 18, 2015

The first (and main) challenge is to factor ${x}^{4} + 1$

Explanation:

${x}^{4} + 1$ has no real zeros, so it has no linear factors with real coefficients.
But fear not, haven't we learned (been told) that every polynomial with real coefficients can be factored into a product of constant, linear and quadratic polynomials with real coefficients? (If we haven't been told that, we should have been.)

We also have need told that every degree n polynomial has n zeros (counting multiplicity). What are the zeros of ${x}^{4} + 1$?

They are the solutions to ${x}^{4} + 1 = 0$
so they solve ${x}^{4} = - 1$
and ${x}^{2} = \pm \sqrt{- 1}$
or ${x}^{2} = \pm i$

So we need the square roots of $i$ and $- i$.

Well, sort of. We know that if $z$ is a solution then so are the conjugate of $z$ and the opposite of $z$, namely $- z$ and the conjugate of $- z$ the is also a solution.

So all we need is just one of the 4 imaginary solutions and we can get the others.

Unless you know something about the geometry of complex numbers (the complex plane) or some version of De Moivre's Theorem, or at least have had a teacher mention that

${\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right)}^{2} = i$, then you may be stuck to find one of the zeros of ${x}^{4} + 1$.

But ${\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right)}^{2} = i$, so it is a zero, as is its conjugate:

$\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i$, its additive inverse: -1/sqrt2 - 1/sqrt2 i),

and the conjugate of that last one: $- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$

Let's use the usual notation $\overline{z}$ for the conjugate.

The conjugate pairs theorem works as it does because

$\left(x - z\right) \left(x - \overline{z}\right)$ is a quadratic with real coefficients.
Lets multiply:

$\left(x - \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right)\right) \left(x - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i\right)\right)$ I'll let you work through it on paper, as I did. But, you'll get:

${x}^{2} - \frac{2}{\sqrt{2}} x + 1 = {x}^{2} - \sqrt{2} x + 1$

Multiply the other pair of factors:
$\left(x - \left(- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i\right)\right) \left(x - \left(- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right)\right)$ to get:

${x}^{2} + \sqrt{2} x + 1$.

Finally, just to convince yourself it worked, multiply it out to verify that

$\left({x}^{2} - \sqrt{2} x + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right) = {x}^{4} + 1$

Now, back to the integral:

For the partial fraction decomposition, you need $A , B , C , D$ to make:

$\frac{A x + B}{{x}^{2} - \sqrt{2} x + 1} + \frac{C x + D}{{x}^{2} + \sqrt{2} x + 1} = \frac{1}{{x}^{4} + 1}$

Have fun!