$$\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(x\right)}{x^4+16}}dx$$
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.
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: $$2I=\ln 4 \int_0^\infty \frac{x^2+4}{x^4+16}dx\Rightarrow I=\ln 2\int_0^\infty \frac{1+\frac{4}{x^2}}{x^2+\frac{4}{x^2}}dx$$ $$=\ln 2\int_0^\infty \frac{d\left(x-\frac{4}{x}\right)}{\left(x-\frac{4}{x}\right)^2+8}=\ln 2 \int_{-\infty}^\infty \frac{dx}{x^2+8}=\frac{\pi}{2\sqrt 2}\ln 2$$
$x=2\tan t,dx=2\sec^2t\ dt$
$$I=\int_0^{\pi/2}\dfrac{8\sec^4t}{16(\tan^4t+1)}\ln(2\tan t)dt$$
$$2I=\int_0^{\pi/2}\dfrac{\ln 2+\ln(\tan t)}{\cos^4t+\sin^4t}dt$$
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 $\tan t=u$
$$I=\int_0^\infty\frac{(4+x^2)\ln x}{16+x^4}\ dx\overset{x=2y}{=}\frac12\int_0^\infty\frac{(1+y^2)(\ln2+\ln y)}{1+y^4}\ dy\\ =\frac{\ln2}{2}\int_0^\infty\frac{1+y^2}{1+y^4}\ dy+\frac12\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy$$
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 $$\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy\overset{y\mapsto 1/y}{=}-\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy\Longrightarrow 2\int_0^\infty\frac{(1+y^2)\ln y}{1+y^4}\ dy=0$$
