Solving $\int_0^{\infty} {1 \over x^4-81}\ dx$?

You can start by getting rid of the constant $81$,

$$\int\frac{dx}{x^4-81}=\frac1{27}\int\frac{dt}{t^4-1}.$$

Then

$$\int\frac{dt}{t^4-1}=\frac12\int\frac1{t^2-1}-\frac12\int\frac1{t^2+1}.$$

The secont term yields $\arctan t$, and the first, if you are aware, $\text{artanh }t$.

Explicitly,

$$\int\frac{dt}{t^2-1}=\frac12\int\frac{dt}{t-1}-\frac12\int\frac{dt}{t+1}=\frac12\log\left|\frac{t-1}{t+1}\right|.$$

---

But, there is a but: the integration range contains a singularity at $t=1$ and the integral is improper.

We can obtain the Cauchy principal value by noting that $\dfrac1{t-1}$ is antisymmetric around $(1,0)$ and we can use

$$\lim_{a\to\infty}\int_0^a\left(\frac1{t-1}-\frac1{t+1}\right)dt=\lim_{a\to\infty}\left(\int_2^a\frac{dt}{t-1}-\int_0^a\frac{dt}{t+1}\right)=\lim_{a\to\infty}(\log a-\log a)=0.$$

It can be done much faster: every well-bred young people should know by heart the following formulæ:

• $\displaystyle\int\frac{\mathrm dx}{a^2+x^2}=\frac1a\,\arctan\Bigl(\frac xa\Bigr)$,
• $\displaystyle\int\frac{\mathrm dx}{a^2-x^2}=\left.\begin{cases}\dfrac1a\,\operatorname{argtanh}\Bigl(\dfrac xa\Bigr)&-a<x<a\\\dfrac1a\,\operatorname{argcoth}\Bigl(\dfrac xa\Bigr)& x<-a \:\text{ or } \:x>a\end{cases} \right\rbrace=\frac1{2a}\,\ln\biggl|\frac{a+ x}{a-x}\biggr|,$ $\quad -a<x<a$.

These formulæ yield instantly $${1\over18}\int_0^\infty\! \Bigl({1 \over x^2-9}-{1 \over x^2+9}\Bigr)\,\mathrm dx=\frac1{18}\biggl[\frac16\ln\biggl|\frac{3-x}{3+x}\biggr|-\frac13\arctan\Bigl(\frac x3\Bigr)\biggr]_0^\infty=\frac1{18}\cdot\frac13\cdot\frac\pi 2$$