Solving $\int_0^{\infty} {1 \over x^4-81}\ dx$? solve the following $$\int_0^{\infty} {dx \over x^4-81}?$$
my approach:
$$\int {dx \over x^4-81}=\int {dx \over (x^2+9)(x^2-9)}$$
used partial fractions
${1\over18}\int ({1 \over x^2-9}-{1 \over x^2+9})dx={1\over 18}\int {dx \over x^2-9}-{1\over 18}\int {dx \over x^2+9}$
for first part, used partial fractions
$${1\over 18}\int_0^{\infty} {1 \over x^2-9}={1\over 108}\int_0^{\infty} ({1 \over x-3}-{1 \over x+3})dx$$$$={1\over 108}\lim_{t\to \infty}\int_0^{t} ({1 \over x-3}-{1 \over x+3})dx$$
$$={1\over 108}\lim_{t\to \infty}(\ln|x-3|-ln|x+3|)_0^{t}$$$$={1\over 108}\lim_{t\to \infty}(\ln|{t-3\over t+3}|)$$ $$={1\over 108}\lim_{t\to \infty}(\ln|{1-3/t\over 1+3/t}|)={1\over 108}\cdot 0=0$$
for second part, I substituted $x=3\tan t$, $dx=3\sec^2 t\ dt$
$${1\over 18}\int_0^{\infty} {dx \over x^2+9}={1\over 18}\int_0^{\pi/2} {3\sec^2t\ dt \over 9\sec^2t}={1\over 54}\int_0^{\pi/2}dt={\pi\over 108} $$
adding two values, my answer becomes $0+({-\pi\over 108})=-{\pi\over 108} $
Can I do this by other way which is simpler than my method? thank you
 A: 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
$$\lim_{a\to\infty}\int_0^a\left(\frac1{t-1}-\frac1{t+1}\right)dt=\lim_{a\to\infty}\left(\int_2^a\frac{dt}{t-1}-\int_0^a\frac{dt}{t+1}\right)=\lim_{a\to\infty}(\log a-\log a)=0.$$
A: 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
$${1\over18}\int_0^\infty\! \Bigl({1 \over x^2-9}-{1 \over x^2+9}\Bigr)\,\mathrm dx=\frac1{18}\biggl[\frac16\ln\biggl|\frac{3-x}{3+x}\biggr|-\frac13\arctan\Bigl(\frac x3\Bigr)\biggr]_0^\infty=\frac1{18}\cdot\frac13\cdot\frac\pi 2$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\mrm{P.V.}\int_{0}^{\infty}
{\dd x \over x^{4} - 81} } =
{1 \over 27}\,\mrm{P.V.}\int_{0}^{\infty}{\dd x \over x^{4} - 1} =
{1 \over 108}\,\mrm{P.V.}\int_{0}^{\infty}{x^{-3/4} \over x - 1}
\,\dd x
\\[5mm] = &\
{1 \over 108}\,\mrm{P.V.}\int_{-1}^{\infty}
{\pars{x + 1}^{-3/4} \over x}\,\dd x
\\[5mm] = &\
{1 \over 108}\int_{0}^{1}{\pars{1 + x}^{-3/4} - \pars{1 - x}^{-3/4} \over x}
\,\dd x + {1 \over 108}\int_{1}^{\infty}{\pars{x + 1}^{-3/4} \over x}\,\dd x
\\[5mm] = &\
\bbx{-\,{\mrm{arccoth}\pars{2^{1/4}} + \arctan\pars{2^{1/4}}\over 54} +
{1 \over 81}\ \mbox{}_{2}\mrm{F}_{1}\pars{{3 \over 4},{3 \over 4};{7 \over 4}\;-1}}
\\[5mm] \approx &\ -0.0291
\end{align}
