how to evaluate the integral $\frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx$ How to evaluate the 59 integral, possibly using real method? 
$$\frac{64}{\pi^3} \int_0^\infty \frac{ (\ln x)^2 (15-2x)}{(x^4+1)(x^2+1)}\ dx$$

Start by evaluating, for $-1<\Re(s)<4$, the following integral:
$$
    I(s) = \int_0^\infty \frac{x^s}{(1+x^2)(1+x^4)} \mathrm{d}x = \frac{1}{2} \int_0^\infty x^s \left( \frac{1}{1+x^2} + \frac{1-x^2}{1+x^4} \right) \mathrm{d}x 
$$
When $-1<\Re(s)<1$ we can rewrite this as a sum of integrals:
$$
   I(s) = \frac{\pi}{4} \frac{1}{\cos\left(\frac{\pi}{2} s\right)} + \frac{\pi}{8} \frac{1}{\cos\left( \frac{\pi}{4} (s+1)\right)} - \frac{\pi}{8} \frac{1}{\cos\left( \frac{\pi}{4} (s+3)\right)}
$$
Where the following result was used:
$$
    \int_0^\infty \frac{x^s}{1+x^n} \mathrm{d} x = \frac{\pi}{n} \frac{1}{\cos\left(  \frac{\pi}{n} (s+1)\right)}
$$
The expression given above remains true for all $-1<Re(s)<4$ by the principle of analytic continuation. Now, we recover the original integral as:
$$
   \frac{64}{\pi^3} \left( 15 I^{\prime\prime}(0) - 2 I^{\prime\prime}(1) \right)
$$
which gives the result.
