$\int_0^\infty \frac{1}{1+x^ 9} \, dx$

There is a simpler way, using Euler's $B$ function:

1) change the variable from $x$ to $y=x^9$. You'll get $\frac 1 9 \int \limits _0 ^\infty \frac {y^{- \frac 8 9}} {1+y} \mathbb{d}y$.

2) make a second change of variables: $t=\frac y {1+y}$; you'll get $\frac 1 9 \int \limits _0 ^1 t^{- \frac 8 9} (1-t)^ {- \frac 1 9} \mathbb{d}t$, which is $\frac 1 9 B(\frac 1 9, \frac 8 9)$.

3) finally, using that $B(x,y)=\frac {\Gamma(x) \Gamma(y)} {\Gamma(x+y)}$ and $\Gamma(x) \Gamma(1-x) = \frac \pi {\sin( \pi x)}$, you'll get $\frac \pi {9 \sin \frac \pi 9}$.

Let's solve a harder problem...$$\int_0^{\infty} {1 \over {1+x^p}} dx$$ Let $u=x^p$... $$\int_0^{\infty} {1 \over {1+x^p}} dx={1 \over p} \int_0^{\infty} {u^{{1 \over p}-1} \over {1+u}} du$$ The definition of the Beta function is... $$B(p,q)=\int_0^{\infty} {{t^{n-1} \over ({t+1})^{p+q}} dt}$$ Note the striking similarity and evaluate accordingly... $${1 \over p} \cdot B \left({1 \over p},1-{1 \over p} \right)$$ Use the Beta function's relation with the Gamma function to get... $${1 \over p} \cdot B \left({1 \over p},1-{1 \over p} \right)={1 \over p} \cdot {{\Gamma \left({1 \over p} \right) \cdot \Gamma \left(1-{1 \over p} \right)} \over {\Gamma (1)}}$$

Use Euler's reflection formula $\Gamma (x) \cdot \Gamma (1-x)= \pi \csc(\pi x)$. You should arrive at... $$\int_0^{\infty} {1 \over {1+x^p}} dx={\pi \over p} \cdot \csc \left( {\pi \over p }\right)$$ For $p=9$... $$\int_0^{\infty} {1 \over {1+x^9}} dx={\pi \over 9} \cdot \csc \left( {\pi \over 9} \right)=1.0206...$$

An easier way to evaluate the general case is to consider a sector surrounding one residue, rather than a whole circle because it's easier to compute the contour integral. To start, consider the function

$$f(z)=\frac 1{z^n+1}$$

Where $n$ is an integer. Now imagine a "pizza slice" contour of radius $R$ pictured below

As $R\to\infty$, the arc integral vanishes. Leaving us with

$$\oint\limits_{C}dz\, f(z)=\int\limits_0^{\infty}dx\, f(x)-e^{2\pi i/n}\int\limits_0^{\infty}dz\, f(z)$$

Here, observe that if you take out a factor of $e^{\pi i/n}$, you get$$e^{\pi i/n}(e^{-\pi i/n}-e^{\pi i/n})\int\limits_0^{\infty}dx\, f(x)=-2i e^{\pi i/n}\sin\left(\frac {\pi}n\right)\int\limits_0^{\infty}dx\, f(x)$$The contour integral as only one residue at $z=e^{\pi i/n}$. The residue is given by

$$\operatorname*{Res}_{z=e^{\pi i/n}}\frac 1{1+z^n}=\lim\limits_{z\to e^{\pi i/n}}\frac {z-e^{\pi i/n}}{z^n+1}=-\frac 1ne^{\pi i/n}$$Therefore$$-2ie^{\pi i/n}\sin\left(\frac {\pi}n\right)\int\limits_0^{\infty}dx\, f(x)=-\frac {2\pi i}ne^{\pi i/n}$$

Simplify both sides you see that

$$\int\limits_0^{\infty}\frac {dx}{1+x^n}=\frac {\pi}n\csc\left(\frac {\pi}n\right)$$