Approach ideas for the integral $\int\frac{dx}{(x^4-16)^2}$

Question

Well, the title sums it up pretty well. I'm in search for some smart approach ideas for solving this indefinite integral: $$\int\frac{dx}{(x^4-16)^2}$$

I know one that would work for sure, namely partial fraction decomposition, but it gets really heavy when regrouping the coefficients for the powers of $x$ and then solving an $8\times8$ system of linear equations. It will eventually work, but I suspect there is something more ingenuine behind this problem.

I also tried all sorts of trigonometric substitutions and formulations, but that added square power really is a bummer to it all.

I'm generally open to any exchange on the topic and would be glad to hear some advice in such situations. Many thanks in advance!

Answer 1

As $x^4-2^4=(x^2-2^2)(x^2+2^2)$ and $(x^2+2^2)-(x^2-2^2)=8$

$$\dfrac{8^2}{(x^4-16)^2}=\dfrac{(x^2+4-(x^2-4))^2}{(x^2-4)^2(x^2+4)^2}=\dfrac1{(x^2-4)^2}+\dfrac1{(x^2+4)^2}-\dfrac2{(x^2-4)(x^2+4)}$$

For the first, write the numerator as $\dfrac{(x+2-(x-2))^2}{16}$ and expand

For the second either $x=2\tan t$

or integrate by parts $$\int\dfrac1x\cdot\dfrac x{(x^2+4)^2}\ dx$$

Again for the last, $\dfrac{(x^2+4)-(x^2-4)}{(x^2-4)(x^2+4)}$

Answer 2

Just integrate by parts

\begin{align} \int\frac{dx}{(x^4-16)^2} & =\int \frac1{64x^3} d\left( \frac{-x^4}{x^4-16}\right) =-\frac1{64} \frac x{x^4-16} -\frac3{64} \int \frac{dx}{x^4-16}\\ &= -\frac1{64} \frac x{x^4-16} -\frac3{64}\cdot\frac18\int \left( \frac1{x^2-4}-\frac1{x^2+4} \right)dx \end{align}

Answer 3

For this particular integral you can use quite a few methods to simplify the computations as shown in the other answers given here. But the more generic partial fraction method can also be used, it's not a tedious method if used correctly. You can avoid solving equations for the coefficients of the terms by expanding around each pole of the integrand. And you can simplify the computations even more by replacing the integrand with repeating factors by one without repeating factors and inserting a parameter in the factors so that you get the correct multiplicity by differentiating w.r.t. the inserted parameters, or by performing a series expansion and extracting an appropriate coefficient from such a series. It's then not necessary to first get to the correct partial fraction expansion and then to compute the integration, the order can be reversed.

In this case we then start with considering the integral:

$$I(a) = \int\frac{dx}{x^4 -a^4} = \frac{1}{a^3}\int\frac{dt}{t^4-1}$$

where $t=\frac{x}{a}$. The poles of the integrand are at $t = \pm 1$ and $t = \pm i $. For a general pole at $t = \alpha$, the singular part of the expansion is:

$$\frac{A(\alpha)}{t-\alpha}$$

where

$$A(\alpha) = \lim_{t\to\alpha} \frac{t-\alpha}{t^4-1} = \frac{1}{4\alpha^3}$$

So, we have the partial fraction expansion:

$$\frac{1}{t^4-1} = \frac{1}{4}\left[\frac{i}{t-i} - \frac{i}{t+i}+\frac{1}{t-1}-\frac{1}{t+1}\right] = \frac{1}{4}\left(\frac{1}{t-1}-\frac{1}{t+1}\right) -\frac{1}{2}\frac{1}{t^2+1}$$

Therefore:

$$I(a) = \frac{1}{a^3}\left[\frac{1}{4}\log\left|\frac{t-1}{t+1}\right|-\frac{1}{2}\arctan(t)\right]$$

We can then write the desired integral as:

$$\int\frac{dx}{\left(x^4 -u\right)^2} = \frac{d I\left(u^{\frac{1}{4}}\right)}{du} = -\frac{3}{4 u^{\frac{7}{4}}}\left[\frac{1}{4}\log\left|\frac{t-1}{t+1}\right|-\frac{1}{2}\arctan(t)\right]-\frac{1}{4 u^{\frac{7}{4}}}\frac{t}{t^4-1} $$

where in the derivative we use that $t = x u^{-\frac{1}{4}}$ and $x$ is kept constant. So, the result:

$$\int\frac{dx}{\left(x^4 -1\right)^2} = \frac{3}{8}\arctan(x)-\frac{1}{4}\frac{x}{x^4-1} -\frac{3}{ 16}\log\left|\frac{x-1}{x+1}\right|$$

can be obtained with very few computations using partial fractions, without having to use special tricks exploiting special properties of the integrand. The only thing that simplified the computations in this case was that the parameter $a$ factors out of the integral, in general this won't happen and the differentiation will yield more terms. However, in such more general cases the answer is then also more complex. In case of real integrands with complex poles, you only need to consider one of each pair of complex conjugates.