<s>[INST] 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.
 [/INST] 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:
$$2I=\ln 4 \int_0^\infty \frac{x^2+4}{x^4+16}dx\Rightarrow I=\ln 2\int_0^\infty \frac{1+\frac{4}{x^2}}{x^2+\frac{4}{x^2}}dx$$
$$=\ln 2\int_0^\infty \frac{d\left(x-\frac{4}{x}\right)}{\left(x-\frac{4}{x}\right)^2+8}=\ln 2 \int_{-\infty}^\infty \frac{dx}{x^2+8}=\frac{\pi}{2\sqrt 2}\ln 2$$
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