I was given a problem to learn how to use Mathematica. I should derive the identity from the paper 1 known as the BBP formula for $\pi$. But I can't figure it out why
$$\begin{equation} \sum_{k=0}^{\infty} \frac{1}{16^k}(\frac{4}{8i+1} - \frac{2}{8i+4}-\frac{1}{8i+5}-\frac{1}{8i+6}) = \int_0^{\frac{1}{\sqrt{2}}} \frac{4\sqrt{2}-8x^3-4\sqrt{2}x^4-8x^5}{1-x^8} {dx} \end{equation}$$
holds while using
$$\begin{equation} \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{k-1}}{1-x^8} {dx} = \int_0^{\frac{1}{\sqrt{2}}} \sum_{i=0}^{\infty} x^{k-1}x^{8i} {dx} = \frac{1}{\sqrt{2}^k} \sum_{i=0}^{\infty} \frac{1}{16i(8i+k)} \end{equation}$$
this. I don't see how I can figure this out neither in Mathematica nor by hand.
Begin by writing $$ \begin{align} \int_0^{\frac{1}{\sqrt{2}}} &\frac{4\sqrt{2}-8x^3-4\sqrt{2}x^4-8x^5}{1-x^8} dx = \\& 4 \sqrt{2}\int_0^{\frac{1}{\sqrt{2}}} \frac{x^0}{1-x^8} dx - 8\int_0^{\frac{1}{\sqrt{2}}} \frac{x^3}{1-x^8} dx \\ & \qquad - 4 \sqrt{2}\int_0^{\frac{1}{\sqrt{2}}} \frac{x^4}{1-x^8} dx - 8\int_0^{\frac{1}{\sqrt{2}}} \frac{x^5}{1-x^8} dx \end{align} $$