How to prove $\sum_{k=0}^{\infty}{1\over 16^k}\left({1\over 8k+1}+{1/2\over 8k+3}-{1/4\over 8k+5}-{1/8\over 8k+7}\right)=\arctan(2)?$ Proposed:

$$\sum_{k=0}^{\infty}{1\over 16^k}\left({1\over 8k+1}+{1/2\over 8k+3}-{1/4\over 8k+5}-{1/8\over 8k+7}\right)=\arctan(2)\tag1$$

My try
Changing $(1)$ into an integral
$$\int_{0}^{1}\left(1+{x^2\over 2}-{x^4\over 4}-{x^6\over 8}\right){\mathrm dx\over 16-x^8}\tag2$$
How may we tackle this integral $(2)?$
 A: Nothing deep, simplify the integrand to
$$
\frac{2+x^2}{32+8x^4}
$$
Then either use a trick, or partial fraction decomposition. You get
$$
\frac1{16}\biggl(\frac{1}{1+(x-1)^2}+\frac{1}{1+(x+1)^2}\biggr)
$$
Integrating, you get (I skip the constant)
$$
\frac{1}{16}\bigl(\arctan(x-1)+\arctan(x+1)\bigr).
$$
In the end, the result is $(1/16)\arctan 2$. I leave it to you to find why you have a factor $1/16$ extra in your integral.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&16\int_{0}^{1}\pars{1 + {x^{2} \over 2} - {x^{4} \over 4} - {x^{6} \over 8}}
{\dd x\over 16-x^{8}}\,\,\,\stackrel{x\ =\ \root{2}{\large t}}{=}\,\,\,
\root{2}\int_{0}^{\root{2}\!/2}{1 + t^{2} \over 1 + t^{4}}\,\dd t
\\[5mm] = &\
\root{2}\int_{0}^{\root{2}\!/2}{1/t^{2} + 1 \over 1/t^{2} + t^{2}}
\,\dd t =
\root{2}\int_{0}^{\root{2}\!/2}{1/t^{2} + 1 \over \pars{1/t - t}^{2} + 2}\,\dd t
\\[5mm] \stackrel{1/t - t \ =\ x}{=}\,\,\,&\
-\root{2}\int_{\infty}^{\root{2}\!/2}{\dd x \over x^{2} + 2} =
\int_{1/2}^{\infty}{\dd x \over x^{2} + 1} = {\pi \over 2} - \arctan\pars{1 \over 2} = \bbx{\arctan\pars{2}}
\end{align}
