Generalization of an Integral Trick? There is an interesting trick that can be used to evaluate integrals in the form
$$I=\int_{-a}^a \frac{E(x)}{b^x+1}dx$$
where $E$ is an even function. Notice that, by substituting $x\to -x$,
$$I=\int_{-a}^a \frac{E(-x)}{b^{-x}+1}dx=\int_{-a}^a \frac{b^xE(x)}{b^{x}+1}dx$$
and so
$$I+I=\int_{-a}^a \frac{E(x)+b^xE(x)}{b^x+1}dx=\int_{-a}^a E(x)dx$$
and so
$$I=\frac{1}{2}\int_{-a}^a E(x)dx=\int_{0}^a E(x)dx$$
For example, this trick can be used to evaluate the intimidating integral
$$\int_{-1}^1 \frac{x^{100}}{e^x+1}dx=\frac{1}{101}$$
QUESTION: Is there some way to generalize this trick to integrals of the form
$$I=\int_{-a}^a \frac{E(x)}{(b^x+1)^2}dx$$
or will this type of integral just have to be done the hard way?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets
  $\ds{{1 \over b^{x} + 1} =
{1 \over \expo{\beta x} + 1} \equiv \mathrm{f}\pars{x}:\ Fermi\!-\!Dirac\ Distribution\ Function.\quad\beta \equiv \ln\pars{b} > 0}$

\begin{align}
&\mbox{It satisfies the identity}\quad
\mrm{f}\pars{x} = 
\left\{\begin{array}{lcrcl}
\ds{\Theta\pars{-x} + \,\mrm{sgn}\pars{x}\mrm{f}\pars{\verts{x}}} & \mbox{if} &
\ds{x} & \ds{\not=} & \ds{0}
\\[2mm]
\ds{1 \over 2} & \mbox{if} & \ds{x} &  \ds{=} & \ds{0}
\end{array}\right.
\\[5mm]
&\mbox{where}\ \Theta\ \mbox{is the}\ Heaviside\ Step\ Function.
\end{align}

Then, with $\ds{a > 0}$:
$$
\bbx{\int_{-a}^{a}\mrm{E}\pars{x}\mrm{f}\pars{x}\dd x =
\int_{-a}^{0}\mrm{E}\pars{x}\dd x + \int_{0}^{a}\bracks{\mrm{E}\pars{x} - \mrm{E}\pars{-x}}\mrm{f}\pars{x}\dd x}
$$


Moreover

\begin{align}
\int_{-a}^{a}{\mrm{E}\pars{x} \over \pars{b^{x} + 1}^{2}}\,\dd x & =
\int_{-a}^{a}\mrm{E}\pars{x}
\bracks{-\,{1 \over \beta}\expo{-\beta x}\mrm{f}'\pars{x}}\,\dd x
\\[1cm] & \stackrel{\mrm{IBP}}{=}\,\,\,
-\,{\mrm{E}\pars{a}\expo{-\beta a}\mrm{f}\pars{a} -
\mrm{E}\pars{-a}\expo{\beta a}\mrm{f}\pars{-a} \over \beta}
\\[2mm] & +
\int_{-a}^{a}\bracks{%
{1 \over \beta}\,\mrm{E}'\pars{x}\expo{-\beta x} -
\mrm{E}\pars{x}\expo{-\beta x}}\mrm{f}\pars{x}\,\dd x
\end{align}

which is reduced to the previous case !!!. 

