Prove ${2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}$ Prove

$$I={2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}\tag1$$

Expand out $(1)$
$$I={2n!\over \pi}\int_{0}^{\infty}{1\over (1+x^2)(2^2+x^2)(3^2+x^2)\cdots(n^2+x^2)}dx\tag2$$
Noticing 
$${1\over (1+x^2)(4+x^2)}={1\over 3}\left({1\over 1+x^2}-{1\over 4+x^2}\right)$$
$${1\over (1+x^2)(4+x^2)(9+x^2)}={1\over 2x^2-12x+30}\left({1\over 1+x^2}-{2\over 4+x^2}+{1\over 9+x^2}\right)$$
$${1\over (1+x^2)(4+x^2)(9+x^2)(16+x^2)}={1\over -60x^2+300}\left({1\over 1+x^2}-{3\over 4+x^2}+{3\over 9+x^2}-{1\over 16+x^2)}\right)$$
and so on...,
k is a polynomial function of x, we have
(I just realised that if k is a function of x, then what follow from (3) are all wrong!)
$${1\over (1+x^2)(2^2+x^2)\cdots(n^2+x^2)}=k\left({{n-1\choose 0}\over 1+x^2}-{{n-1\choose 1}\over 2^2+x^2}+{{n-1\choose 2}\over 3^2+x^2}-\cdots{{n-1\choose n-1}\over n^2+x^2}\right)\tag3$$
Recall
$$\int_{0}^{\infty}{1\over a^2+x^2}dx={\pi\over 2a}\tag4$$
Sub $(3)$ into $(2)$ and applying $(4)$ hence we have
$$I={2n!k\over \pi}\cdot{\pi\over 2}\left[{{n-1\choose 0}\over 1}-{{n-1\choose 1}\over 2}+{{n-1\choose 2}\over 3}-\cdots-{{n-1\choose n-1}\over n}\right]\tag5$$
$$I={n!k}\left[{{n-1\choose 0}\over 1}-{{n-1\choose 1}\over 2}+{{n-1\choose 2}\over 3}-\cdots-{{n-1\choose n-1}\over n}\right]\tag6$$
How can we get from $(6)$ to $I={n\over 2n-1}$?
Can anyone produce another method less lengthy than this method above to tackle Integral (1)?
I have saw some authors using the residue theorem to tackle another simple case like the above (1) but I don't know how to apply it.
 A: You already did the $95\%$ of the needed work through partial fraction decomposition. You just need to compute:
$$ J(n) = \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{(-1)^k}{k+1} = \int_{0}^{1}\sum_{k=0}^{n-1}\binom{n-1}{k}(-x)^k\,dx $$
that is:
$$ J(n) = \int_{0}^{1}(1-x)^{n-1}\,dx = \int_{0}^{1} x^{n-1}\,dx = \frac{1}{n},$$
no big issue. Restarting from scratch,
$$ \int_{-\infty}^{+\infty}\frac{dx}{f_n(x)} = \int_{-\infty}^{+\infty}\prod_{k=1}^{n}\frac{1}{1+\frac{x^2}{k^2}}\,dx =2\pi i\cdot\!\!\!\!\sum_{\substack{z\in[-k,k]\\ z\neq 0}}\!\!\text{Res}\left(\frac{1}{f_n(x)},x=zi\right)\tag{1}$$
but since the singularities of $f_n(x)$ are just simple poles, De L'Hopital theorem gives:
$$ \sum_{\substack{z\in[-k,k]\\ z\neq 0}}\!\!\text{Res}\left(f_n(x),x=zi\right)=\sum_{\substack{z\in[-k,k]\\ z\neq 0}}\frac{1}{f_n'(zi)}\tag{2}$$
and $f_n(x)$ is a product, to it looks like a good idea to use logarithmic differentiation:
$$ f_n'(x) = f_n(x)\cdot\frac{d}{dx}\log f_n(x) = \left(\sum_{h=1}^{n}\frac{2h}{x^2+h^2}\right)\prod_{h=1}^{n}\left(1+\frac{x^2}{h^2}\right)\tag{3}$$
that gives the wanted partial fraction decomposition. It is also interesting to point out that this exercise gives an unusual proof of:
$$ \int_{0}^{+\infty}\frac{x}{\sinh x}\,dx = \frac{\pi^2}{4} \tag{4}$$
since the Weierstrass product for the $\sinh $ function is uniformly convergent over any compact subset of $\mathbb{R}$.
A: I dont know if the follow help you. put  $$I_n={2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n}{k^2\over k^2+x^2}dx$$ By induction we have: 
$$I_1={2\over \pi}\int_{0}^{\infty}{1\over 1+x^2}dx={2\over \pi}{ \pi\over2}=1$$
Suppose that it true for $I_n$. Now
\begin{align}I_{n+1}&={2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{k^2\over k^2+x^2}\right){(n+1)^2\over (n+1)^2+x^2}dx\\
&={2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{k^2\over k^2+x^2}\right)dx-{2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{k^2\over k^2+x^2}\right){x^2\over (n+1)^2+x^2}dx\\
&=I_n-{2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{(kx)^2\over k^2+x^2}\right){1\over (n+1)^2+x^2}dx
\end{align}
and so we are able to finish the proof if we can prove that
$$\int_{0}^{\infty}\left(\prod_{k=1}^{n}{(kx)^2\over k^2+x^2}\right){1\over (n+1)^2+x^2}dx={\pi\over 2(2n-1)(2n+1)}$$
