Call it what you want

$\displaystyle f(n)=\int_0^\infty\frac{1}{1+x^n}dx=\int_{0}^{\infty}\int_{0}^{\infty}e^{-t(1+x^n)}dt\ dx$

Sub $\displaystyle u = x^nt\quad du = nx^{n-1}t\,dx$

$\displaystyle\frac{1}{n}\int_{0}^{\infty}e^{-t}t^{\frac{-1}{n}}dt\int_{0}^{\infty}u^{\frac{1}{n}-1}e^{-u}du =$

$\displaystyle\frac{\Gamma(1-\frac{1}{n})\Gamma(\frac{1}{n})}{n} = \frac{\pi}{n\sin{\frac{\pi}{n}}}\quad \text{by Euler's Reflection Formula}$

$\displaystyle\int_2^{10}f^2(n)dn=\int_0^{10}\frac{\pi^2}{n^2\sin^2{\frac{\pi}{n}}}dn=$

$\displaystyle -\pi^2\int_\frac{1}{2}^\frac{1}{10}\csc^2{\pi x}dx =\pi\cot{\frac{\pi}{10}}=\pi\sqrt{5+2\sqrt{5}}$

I have a different, much lengthier approach :(

$f(n)=\int_{0}^{\infty} \frac{dx}{1+x^{n}}$

Let $t=\frac{1}{1+x^{n}}$

$dt=\frac{-nx^{n-1}}{{1+x^{n}}^2}dx=-nt^{2}x^{n-1}dx$

$t=\frac{1}{1+x^{n}}$ , $\therefore x^{n-1}=(\frac{1-t}{t})^{\frac{n-1}{n}}$

$\therefore dt=-nt^{2}(\frac{1-t}{t})^{\frac{n-1}{n}}dx$

Rearranging,

$dx=-\frac{1}{n}t^{-2}(\frac{1-t}{t})^{\frac{1-n}{n}}dt$

Using this and changing the limits,

$f(n)=\int_{1}^{0}-\frac{t}{n}t^{-2}(\frac{1-t}{t})^{\frac{1-n}{n}}dt =\int_{0}^{1} \frac{t}{n}t^{-2} t^{\frac{1-n}{n}}({1-t})^{\frac{1-n}{n}}dt$

Algebra...

$\therefore f(n)=\frac{1}{n}\int_{0}^{1} t^{(1-\frac{1}{n})-1}(1-t)^{\frac{1}{n}-1}= B(1-\frac{1}{n},\frac{1}{n})$

where $B(x,y)$ is the beta function, with these properties:

$B(x,y)=\int_{0}^{1} t^{x-1}(1-t)^{y-1}dt$

$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

The first one was used in the integral. Using the second one,

$f(n)=\frac{\Gamma(\frac{1}{n})\Gamma(1-\frac{1}{n})}{n\Gamma(1)}$ which using Euler's reflection formula and $\Gamma(n)=(n-1)!$ becomes

$f(n)=\frac{\pi}{n\sin(\frac{\pi}{n})} =\frac{\pi}{n}\csc(\frac{\pi}{n})$

$\int_{2}^{10}\frac{f^{2}(n)}{\pi}dn=\int_{2}^{10} \frac{\pi}{n^{2}}\csc(\frac{\pi}{n})dn$

Let $u=\frac{\pi}{n}$ , $du=-\frac{\pi}{n^{2}}dn$

$\int_{2}^{10}\frac{f^{2}(n)}{\pi}dn=\int_{\frac{\pi}{2}}^{\frac{\pi}{10}} -\csc^{2}(x)dx = \cot(x) \bigg\vert_{\frac{\pi}{2}}^{\frac{\pi}{10}} = \cot(\frac{\pi}{10})$

$\cot(\frac{\pi}{10})$ :

1. find $\cos(\frac{\pi}{10})$ by expanding and solving the fifth-degree polynomial from $\cos(5\theta)=0$ when $\theta=\frac{\pi}{10}$ (there has to be a better way)
2. use $\cos\theta$ to get $\sin\theta$
3. $\cot\theta=\frac{\cos\theta}{\sin\theta}$
4. $\therefore \cot(\frac{\pi}{10}) = \sqrt{5+2\sqrt{5}}$

$\therefore \int_{2}^{10}\frac{f^{2}(n)}{\pi}dn=\sqrt{5+2\sqrt{5}}$

$2+5=7$