Interesting Integral

$x^4 +16 = (x^2 - 2\sqrt{2} x +4)(x^2+2\sqrt{2} x + 4)$

Hence,

$I = \displaystyle \int_{0}^{\infty} \dfrac{1}{2} \bigg(\dfrac{\ln x }{x^2-2 \sqrt{2} x+ 4} + \dfrac{\ln x}{x^2+ 2\sqrt{2} x+4} \bigg) {\mathrm dx}$

Now, Putting $x = 2z$ ,

$\displaystyle I = \dfrac{1}{4} \displaystyle \int_{0}^{\infty} \bigg(\dfrac{\ln 2 + \ln z}{z^2-\sqrt{2} z +1} + \dfrac{\ln 2 + \ln z}{z^2+\sqrt{2} z +1} \bigg) {\mathrm dz}$

Now, putting $z = \dfrac{1}{t}$ ,

$\displaystyle I = \dfrac{1}{4} \displaystyle \int_{0}^{\infty} \bigg(\dfrac{\ln 2 - \ln t}{t^2-\sqrt{2} t +1} + \dfrac{\ln 2 - \ln t}{t^2+\sqrt{2} t +1} \bigg) {\mathrm dt}$

Adding these, we get :

$I = \displaystyle \int_{0}^{\infty} \dfrac{\ln 2}{4}\bigg( \int_{0}^{\infty} \dfrac{1}{z^2 - \sqrt{2} z + 1} {\mathrm dz} +\int_{0}^{\infty} \dfrac{1}{z^2+\sqrt{2} z +1} {\mathrm dz} \bigg)$

= $\dfrac{\ln 2}{4} \bigg(\dfrac{3 \pi}{2 \sqrt{2}} + \dfrac{\pi}{2 \sqrt{2}}\bigg)$

= $\boxed{\dfrac{\pi \ln 2}{2 \sqrt{2}}}$

Let's start from the formula I derived here : $\int_0^\infty\dfrac{x^{\,\large a-1}}{1+x^{\,\large b}}\ dx=\frac{\pi}{b}\csc\left(\frac{a\pi}{b}\right)\qquad\qquad\qquad\tag{1}$ Setting $x=\frac{t}{c}$ to $(1)$ , we have $\int_0^\infty\dfrac{t^{\,\large a-1}}{c^{\,\large b}+t^{\,\large b}}\ dt=\frac{\pi}{b\,c^{\,b-a}}\csc\left(\frac{a\pi}{b}\right)\qquad\qquad\qquad\tag{2}$ Differentiating $(2)$ with respect to $a$ yields $\int_0^\infty\dfrac{t^{\,\large a-1}\ln(t)}{c^{\,\large b}+t^{\,\large b}}\ dt=\frac{\partial }{\partial a}\left[\frac{\pi}{b\,c^{\,b-a}}\csc\left(\frac{a\pi}{b}\right)\right]\qquad\qquad\qquad\tag{3}$ Using $(3)$ to solve the considered integral by setting $b=4$ , $c=2$ , and making substitution $t\mapsto x$ , then $\begin{aligned}\int_0^\infty\dfrac{\left(x^{2}+4\right)\ln x}{x^4+16}\ dt&=\frac{\partial }{\partial a}\left[\frac{\pi}{4\cdot2^{\,4-a}}\csc\left(\frac{a\pi}{4}\right)\right]_{a=3}+4\frac{\partial }{\partial a}\left[\frac{\pi}{4\cdot2^{\,4-a}}\csc\left(\frac{a\pi}{4}\right)\right]_{a=1}\\&=\color{#3D99F6}{\frac{\pi\ln2}{2\sqrt{2}}}\end{aligned}$

Put $x=2t$ to get our integral as :

$\displaystyle I=\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \frac { { (t }^{ 2 }+1)ln(2t) }{ ({ t }^{ 4 }+1) } dt }$

$\displaystyle I=\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \frac { { (t }^{ 2 }+1)ln(t) }{ ({ t }^{ 4 }+1) } dt } +\frac { ln(2) }{ 2 } \int _{ 0 }^{ \infty }{ \frac { { (t }^{ 2 }+1) }{ ({ t }^{ 4 }+1) } dt }$

$I={I}_{1}+{I}_{2}$

$\displaystyle {I}_{1}=\int _{ 0 }^{ 1 }{ \frac { { (t }^{ 2 }+1)ln(t) }{ ({ t }^{ 4 }+1) } dt } +\int _{ 1 }^{ \infty }{ \frac { { (t }^{ 2 }+1)ln(t) }{ ({ t }^{ 4 }+1) } dt }$

In the second integral put $z=1/t$ to get :

$\displaystyle {I}_{1}=\int _{ 0 }^{ 1 }{ \frac { { (t }^{ 2 }+1)ln(t) }{ ({ t }^{ 4 }+1) } dt } -\int _{ 0 }^{ 1 }{ \frac { { (z }^{ 2 }+1)ln(z) }{ (z^{ 4 }+1) } dz }=0$

$\displaystyle {I}_{2}=\frac { ln(2) }{ 2 } \int _{ 0 }^{ \infty }{ \frac { { (t }^{ 2 }+1) }{ ({ t }^{ 4 }+1) } dt }=\frac { ln(2) }{ 2 } \int _{ 0 }^{ \infty }{ \frac { (1+\frac { 1 }{ { t }^{ 2 } } ) }{ ({ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } -2)+2 } dt }$

Put $t-\frac{1}{t}=x$ to get ${I}_{2}$ as :

$\displaystyle {I}_{2}=\frac { ln(2) }{ 2 } \int _{ 0 }^{ \infty }{ \frac { dx }{ { x }^{ 2 }+2 } }$

$\Rightarrow {I}_{2}=\frac { \pi ln(2) }{ 2\sqrt { 2 } }$

Hence $I=\frac { \pi ln(2) }{ 2\sqrt { 2 } }$

Another approach........... Put x^2 = 4tan(theta).............then, differentiate under the integral sign..........!!!!