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Relevant wiki: Integration Tricks

We can factor the denominator as $(\frac{1}{2}x^{2}+1)^{2}$ , and there is an $x dx$ in the numerator, so it looks like a good substitution to make would be to let $u=\frac{1}{2}x^{2}$ . With this substitution, we can see that the bounds of integration will stay the same.

$I=\int_{0}^{\infty} \frac{\ln{(\sqrt{32u})}}{(u+1)^{2}} du = \frac{1}{2} \int_{0}^{\infty} \frac{\ln{(32u)}}{(u+1)^{2}}$

For any function $f(x)$ that is continuous and differentiable over the interval $(0,\infty)$ , we can generalize the following: $R=\int_{0}^{\infty}f(x) dx$ . Let $x=\frac{1}{y}$ , then $dx=-\frac{1}{y^{2}}dy$ . Also, the bounds of integration will flip, since $\lim_{x \rightarrow 0} y(x)=\infty$ , and $\lim_{x \rightarrow \infty} y(x)=0$ . Thus, we get that $R=\int_{\infty}^{0} -\frac{f(\frac{1}{y})}{y^{2}}dy=\int_{0}^{\infty}\frac{f(\frac{1}{y})}{y^{2}}dy$ .

We can now rewrite $R$ by adding the two integrals we have written for R, and then dividing by two. Hence, $R=\frac{1}{2} \int_{0}^{\infty} f(x)+\frac{f(\frac{1}{x})}{x^{2}} dx$ .

Applying this to our original integral:

$I=\frac{1}{4} \int_{0}^{\infty} \frac{\ln{(32u)}}{(u+1)^{2}} + \frac{\ln{(\frac{32}{u})}}{(\frac{1}{u}+1)^{2}u^{2}} du = \frac{1}{4} \int_{0}^{\infty} \frac{\ln{(32u)}+\ln{(\frac{32}{u})}}{(u+1)^{2}} du$

This nicely condenses into:

$I=\frac{5 \ln{(2)}}{2} \int_{0}^{\infty} \frac{1}{(u+1)^{2}} du=\frac{5 \ln{(2)}}{2} \approx 1.7328...$

And lastly, $\lfloor 100I \rfloor = \boxed{173}$

the u sub $\begin{cases} u=\dfrac{1}{\frac{1}{2}x^2+1} \implies x^2=2\frac{1-u}{u}\\ |_{0}^{\infty} \to |_1^0\\ du=\dfrac{-x}{(\frac{1}{2}x^2+1)^2} dx \end{cases}$ the integral becomes

$\int_0^1 \ln(4x) du=\int_0^1 \left(\ln(4)+\frac{1}{2}\left(\ln(2)+\ln(1-u)-\ln(u)\right)\right) du\\ =2.5\ln(2)+.5\int_0^1(\ln(1-u)-\ln(u))du$ by subbing 1-u to the int, we will get it = to its negative and hence 0. so the answer is $\boxed{2.5\ln(2)}$

Looking back on it I could have done this with one substitution but as a novice mathematician, it was necessary for me to break down the problem first by reducing from quartic to quadratic before performing the inversion technique. Furthermore to ascertain that the inversion of the variable should be $\frac{4}{v}$ in the following solution, I applied this method:

Let substitution be $u = \frac{k}{v}$ , $du = -\frac{k}{v^{2}}$

Before the substitution, the denominator is $(u+2)^{2}$ , and afterwards it will become $-v^{2}(\frac{k}{v}+2)^{2} = (2v+k)^{2}$ . Since these two must be in the same ratio for the final step to work (by adding numerators over the same denominator), clearly $k = 4$ . Now, on to the integration:

$\\I = {\displaystyle \int_{0}^{\infty}}\frac{x \ln (4x)}{\frac{1}{4}x^{4}+x^{2}+1}dx = 4{\displaystyle \int_{0}^{\infty}}\frac{x \ln (4x)}{x^{4}+4x^{2}+4}dx = 4{\displaystyle \int_{0}^{\infty}}\frac{x \ln (4x)}{(x^{2}+2)^{2}}dx$

Let $u = x^{2}$ , $dx = \frac{du}{2x}$

$I = 2{\displaystyle \int_{0}^{\infty}}\frac{\ln (4u^{\frac{1}{2}})}{(u+2)^{2}}du = 2{\displaystyle \int_{0}^{\infty}}\frac{2 \ln (2) + \frac{1}{2} \ln (u)}{(u+2)^{2}}du\quad$ (1)

Let $u = \frac{4}{v}$ , $du = -\frac{4}{v^{2}}$

$I = -8{\displaystyle \int_{\infty}^{0}}\frac{2 \ln (2) + \frac{1}{2} \ln (\frac{4}{v})}{v^{2}(\frac{4}{v}+2)^{2}}dv = 8{\displaystyle \int_{0}^{\infty}}\frac{3 \ln (2) - \frac{1}{2} \ln (v)}{(4+2v)^{2}}dv = 2{\displaystyle \int_{0}^{\infty}}\frac{3 \ln (2) - \frac{1}{2} \ln (v)}{(v+2)^{2}}dv \quad$ (2)

Now using the equivalence of (1) and (2) ...

$I = \frac{1}{2}\bigg(2{\displaystyle \int_{0}^{\infty}}\frac{2 \ln (2) + \frac{1}{2} \ln (u)}{(u+2)^{2}}du + 2{\displaystyle \int_{0}^{\infty}}\frac{3 \ln (2) - \frac{1}{2} \ln (v)}{(v+2)^{2}}dv \bigg) = {\displaystyle \int_{0}^{\infty}}\frac{5 \ln (2)}{(u+2)^{2}}du = -\left[\frac{5 \ln(2)}{u+2}\right]_0^\infty = \frac{5}{2} \ln(2)$

$\left \lfloor{100I}\right \rfloor = \left \lfloor{250 \ln(2)}\right \rfloor = \boxed{173}$