Partial Fraction Decomposition trouble $\frac{1}{(1+x^2)(1+x^{2015})}$ I wish to decompose $\frac{1}{(1+x^2)(1+x^{2015})}$
I have $1 = (Ax+B)(1+x^{2015}) + (Cx+D)(1+x^2) = Ax^{2016} + Bx^{2015} + Cx^{3} + Dx^{2} + x(A+C) + B+D$
Doesn't this imply that $B+D = 1$, but we also have that $B=D = 0$ as well.
What did I do wrong?

I am examining the integral $\int_{0}^{\infty} \frac{dx}{(1+x^2)(1+x^{2015})}$

There is a trick:
$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}&=&\int_{0}^{1}\frac{dx}{(1+x^2)(1+x^m)}+\int_{1}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}\\&=&\int_{0}^{1}\frac{dx}{1+x^2}\left(\frac{1}{1+x^m}+\frac{1}{1+x^{-m}}\right)\\&=&\int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\color{red}{\frac{\pi}{4}}\end{eqnarray*}$$
no matter what $m$ is.
Explanation: split the integration range into $(0,1)\cup (1,+\infty)$; use the change of variable $x\mapsto\frac{1}{x}$ on the unbounded piece and recombine the two pieces in a single integral over $(0,1)$; simplify, simplify more.
