Evaluate $ \int_0^1 \sum_{k=0}^\infty (-x^4)^k dx = \int_0^1 \frac{dx}{1+x^4} $ I have read this thread and I found in some comments the above named equality. I couldn't follow the transformation, which are done to get from the left to the right side at that point in particular.
Can someone help me and show how it's done? 

On the left, reverse the order of summation and integration to get
$$\sum_{k=0}^{\infty} (-1)^k \, \int_0^1 dx \, x^{4 k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+1} = 1 - \frac15 + \frac19 - \frac1{13} + \cdots$$
This sum is equal to 
$$\int_0^1 \frac{dx}{1+x^4}$$
To evaluate this, observe that
$$1+x^4 = (1+\sqrt{2} x + x^2)(1-\sqrt{2} x+x^2)$$
so that we may use partial fractions in the integrand as follows:
$$\frac1{1+x^4} = \frac{A x+B}{x^2-\sqrt{2} x+1} + \frac{C x+D}{x^2+\sqrt{2} x+1}$$
where one may deduce that $A=-1/(2 \sqrt{2})$, $B=1/2$, $C=1/(2 \sqrt{2})$, and $D=1/2$.  We rewrite the resulting expression for ease of integration:
$$\begin{align}\int_0^1 \frac{dx}{1+x^4} &= \frac1{2} \int_0^1 dx \, \left (\frac{-\frac1{\sqrt{2}} \left (x-\frac1{\sqrt{2}} \right )+\frac12}{\left (x-\frac1{\sqrt{2}} \right )^2+\frac12} + \frac{\frac1{\sqrt{2}} \left (x+\frac1{\sqrt{2}} \right )+\frac12}{\left (x+\frac1{\sqrt{2}} \right )^2+\frac12} \right ) \\ &= \frac1{4 \sqrt{2}} \left [\log{\left (\frac{1+\frac1{\sqrt{2}}}{1-\frac1{\sqrt{2}}} \right )} + 2 \arctan{(\sqrt{2}-1)} + 2 \arctan{(\sqrt{2}+1)} \right ] \\ &=  \frac1{4 \sqrt{2}} \left [\log{\left (\frac{1+\frac1{\sqrt{2}}}{1-\frac1{\sqrt{2}}} \right )} + \pi\right ]\end{align}$$
Thus,
$$\begin{align}\sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+1} = 1 - \frac15 + \frac19 - \frac1{13} + \cdots &= \frac1{4 \sqrt{2}} \left [\pi+\log{\left (\frac{1+\frac1{\sqrt{2}}}{1-\frac1{\sqrt{2}}} \right )} \right ]\\ &=\frac1{4 \sqrt{2}} \left [\pi + 2 \log{\left (\sqrt{2}+1\right )} \right ]\end{align}$$
ADDENDUM
I may as well illustrate how to get the sum of another, similar series almost for free from the above series.  Note that
$$\int_1^{\infty} \frac{dx}{1+x^4} = \int_0^1 dx \frac{x^2}{1+x^4} = \sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+3} $$
However,
$$\begin{align}\int_1^{\infty} \frac{dx}{1+x^4} &= \int_0^{\infty} \frac{dx}{1+x^4} -\int_0^{1} \frac{dx}{1+x^4} \end{align} $$
The first integral on the right may be evaluated using a contour integral and the residue theorem.  I prefer to use a quarter-circle in the upper-right quadrant; the only enclosed pole is at $z=e^{i \pi/4}$.  By the residue theorem, we have
$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4} = \frac{i 2 \pi}{4 e^{i 3 \pi/4}} \implies \int_0^{\infty} \frac{dx}{1+x^4} = \frac{\pi}{2 \sqrt{2}}$$
Thus,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{4 k+3} = \frac13 - \frac17 + \frac1{11} - \frac1{15} + \cdots = \frac1{4 \sqrt{2}} \left [\pi - 2 \log{\left (\sqrt{2}+1\right )} \right ] $$
