Evaluating $\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x$ I need to evaluate $$\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots  +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x
$$
I've been told that the way forward is showing that the integral is the same as $$\int_0^{\infty} (x^{2016} + 1)^{-1} \, \mathrm{d}x$$
i.e: that the weird sum of fractions doesn't affect the integral. 
I've tried $$\sum_{n=1}^{2015} \frac{n}{n+x} = \sum_{n=1}^{2015} \left(1 - \frac{x}{n+x}\right) = 2015 - \sum_{n=1}^{2015} \frac{x}{n+x}$$
but it's getting me nowhere. 
 A: Edited: This integral evaluation is of a related integral which is equivalent to $\int^{\infty}_0(x^{2016}+1)^{-1}dx$, for the original integral in the question, differs in evaluation for limits from $0$ to $\infty$ owing to symmetry considerations discussed in the comments.

$$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots  +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x
$$
$$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x
$$
Now, letting $f(x)=\frac{1}{x^{2016}+1}$, and noting that $f(x)=f(-x)$,
$$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\mathrm{d}x=\frac{1}{2}\int_{-\infty}^{\infty} f\left(-\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\right)\mathrm{d}x$$
$$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(x-\sum^{2015}_{i=1}\frac{i}{x-(-i)}\right)\mathrm{d}x \tag {1}$$
Using Glasser's Master Theorem, 
$$I=\frac{1}{2}\int^{\infty}_{-\infty} f(x)\ \mathrm{d}x=\frac{1}{2}\int^{\infty}_{-\infty} \frac{1}{x^{2016}+1}\ \mathrm{d}x=\int^{\infty}_{0} \frac{1}{x^{2016}+1}\ \mathrm{d}x \tag {2}$$ 

Now we know that $$B(a,b)=\int^{\infty}_0\frac{t^{a-1}}{(1+t)^{a+b}}dt$$
From $(2)$,after substituting $x^{2016} =t$,
$$I=\frac{1}{2016}\int^{\infty}_{0}\frac{t^{\frac{1}{2016}-1}}{(1+t)^{\frac{1}{2016}+\frac{2015}{2016}}}dt=\frac{1}{2016}B(\frac{1}{2016},\frac{2015}{2016})$$
Therefore $$\color{red}{I=\frac{1}{2016}\frac{\Gamma(\frac{1}{2016})\Gamma(\frac{2015}{2016})}{\Gamma(1)}=\frac{\pi}{2016\sin(\frac{\pi}{2016})}\approx1.0000004047320180811575}$$
