Integrate $\int_{0}^{\infty}{\frac{(x^2+4)\ln(x)}{x^4+16}}dx$ 
$$\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(x\right)}{x^4+16}}dx$$

I tried indefinite integration and got struck at $$\int_{0}^{\infty}{\dfrac{\left(u^2+1\right)\ln\left(u\right)}{u^4+1}}du.$$ 
I tried by-parts now taking $\ln(u)$ as my first function, but it doesn't seem to help. How to solve it and also is there an easy way out to solve it, as my method isn't that good.
 A: Start by letting $x\to \frac{4}{x}$ then you get:
$$I=\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(x\right)}{x^4+16}}dx=\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(\frac{4}{x}\right)}{x^4+16}}dx$$
Now add them up to get:
$$2I=\ln 4 \int_0^\infty \frac{x^2+4}{x^4+16}dx\Rightarrow I=\ln 2\int_0^\infty \frac{1+\frac{4}{x^2}}{x^2+\frac{4}{x^2}}dx$$
$$=\ln 2\int_0^\infty \frac{d\left(x-\frac{4}{x}\right)}{\left(x-\frac{4}{x}\right)^2+8}=\ln 2 \int_{-\infty}^\infty \frac{dx}{x^2+8}=\frac{\pi}{2\sqrt 2}\ln 2$$
A: $x=2\tan t,dx=2\sec^2t\ dt$
$$I=\int_0^{\pi/2}\dfrac{8\sec^4t}{16(\tan^4t+1)}\ln(2\tan t)dt$$
$$2I=\int_0^{\pi/2}\dfrac{\ln 2+\ln(\tan t)}{\cos^4t+\sin^4t}dt$$
Like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.,
$$I=\ln2\int_0^{\pi/2}\dfrac{dt}{\cos^4t+\sin^4t}$$
Now divide numerator and denominator by $\cos^4t$ and choose $\tan t=u$
A: $$I=\int_0^\infty\frac{(4+x^2)\ln x}{16+x^4}\ dx\overset{x=2y}{=}\frac12\int_0^\infty\frac{(1+y^2)(\ln2+\ln y)}{1+y^4}\ dy\\
=\frac{\ln2}{2}\int_0^\infty\frac{1+y^2}{1+y^4}\ dy+\frac12\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy$$
where 
$$\int_0^\infty\frac{1+y^2}{1+y^4}\ dy=\int_0^\infty\frac{1/y^2+1}{1/y^2+y^2}\ dy=\int_0^\infty\frac{1+1/y^2}{(y-1/y)^2+2}\ dy\\=\left.\frac1{\sqrt{2}}\tan^{-1}\frac{(y-1/y)}{\sqrt{2}}\right|_0^\infty=\frac{\frac{\pi}{2}-(-\frac{\pi}{2})}{\sqrt{2}}=\frac{\pi}{\sqrt{2}}$$
and $$\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy\overset{y\mapsto 1/y}{=}-\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy\Longrightarrow 2\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy=0$$
