$\int$

Putting $y = x^n$ we see that $\begin{aligned} \int_0^\infty \frac{\ln x}{1 + x^n}\,dx & = \; \frac{1}{n^2} \int_0^\infty \frac{y^{\frac{1}{n}-1}\,\ln y}{1+y}\,dy \; = \; \frac{1}{n^2} \frac{\partial}{\partial \alpha} B(\alpha,1-\alpha) \Big|_{\alpha = \frac{1}{n}} \\ & = \; \frac{1}{n^2} \frac{\partial}{\partial \alpha} \Big[\Gamma(\alpha)\Gamma(1-\alpha)\Big]\Big|_{\alpha = \frac{1}{n}} \; = \; \frac{1}{n^2} \frac{\partial}{\partial \alpha} \big[\pi \mathrm{cosec}\,\pi\alpha\big]\Big|_{\alpha = \frac{1}{n}} \\ & = \; -\tfrac{\pi^2}{n^2} \mathrm{cosec}\tfrac{\pi}{n}\,\cot\tfrac{\pi}{n} \end{aligned}$ Thus $\int_0^\infty \frac{\ln x}{1 + x^{\frac32}}\,dx \; = \; \tfrac{8}{27}\pi^2 \hspace{2cm} \int_0^\infty \frac{\ln x}{1 + x^3}\,dx \; = \; -\tfrac{2}{27}\pi^2$ making the answer $\boxed{-4}$ .

$\text{let } =g(n)=2n \\ (f\circ g)(n)=\int_0^\infty{\dfrac{\ln x}{1+x^{2n}} dx} \\$

$\text{using Complex Integration } \\ \oint {\frac{\ln(z)}{1+z^{2n}}dz} =\int _{ \Gamma_R }{\frac{\ln(z)}{1+z^{2n}}dz} +\int _{-R}^{-r}{\frac{\ln(z)}{1+z^{2n}}dz} +\int _{ \Gamma_r }{\frac{\ln(z)}{1+z^{2n}}dz} +\int _{r}^{R}{\frac{\ln(z)}{1+z^{2n}}dz} \\ \begin{matrix} \color{#D61F06}\text{as }r\rightarrow 0 \text{ and } R \rightarrow \infty \Rightarrow \int _{ \Gamma_R }{\frac{\ln(z)}{1+z^{2n}}dz}=0 \text{ and } \int _{ \Gamma_r }{\frac{\ln(z)}{1+z^{2n}}dz} & & ML-inequality \end{matrix} \\ \begin{matrix} \color{#3D99F6} \int _{-R}^{-r}{\frac{\ln(z)}{1+z^{2n}}dz}=\int _{-R}^{-r}{\frac{\ln(-z)+i\pi}{1+z^{2n}}dz}=\int _{r}^{R}{\frac{\ln(z)+i\pi}{1+z^{2n}}dz}& & & z<0\Rightarrow \ln(-z)=ln\left | z \right |+i\pi \end{matrix} \\ \Rightarrow \oint {\frac{\ln(z)}{1+z^{2n}}dz} =2\int _{0}^{\infty}{\frac{\ln(x)}{1+x^{2n}}dx} +i\pi \int _{0}^{\infty}{\frac{1}{1+x^{2n}}dx} \rightarrow (*)$

$\color{#69047E} {\text{let z}=e^{i\theta} \\ \Rightarrow z^{2n}+1=0\Leftrightarrow e^{2i n\theta}+1=0\Leftrightarrow \theta=\frac{(2k+1)\pi}{2n} \text{ for }k=0,1,2,\dots ,n-1 \\ Res\left ( \frac{\ln(z)}{1+z^{2n}},z_k \right )=\lim _{ z\rightarrow z_k }{ \frac{\ln(z)}{1+z^{2n}} \cdot (z-z_k) } =\frac{z_k\ln(z_k)}{2n{z_k}^{2n}}=\frac{1}{2ne^{i(2k+1)\pi}} \cdot e^{\frac{i(2k+1)\pi}{2n}}\cdot \frac{i(2k+1)\pi}{2n}=\frac{-i\pi}{4n^2}(2k+1)\cdot e^{\frac{i(2k+1)\pi}{2n}} \\ \oint {\frac{\ln(z)}{1+z^{2n}}dz}=2\pi i \sum_{k=0}^{n-1}{Res\left ( \frac{\ln(z)}{1+z^{2n}},z_k \right )}=2\pi i\sum_{k=0}^{n-1}\frac{-i\pi}{4n^2}(2k+1)\cdot e^{\frac{i(2k+1)\pi}{2n}} \\ \oint {\frac{\ln(z)}{1+z^{2n}}dz}=\frac{\pi^2}{2n^2}\sum_{k=0}^{n-1}{(2k+1)\left ( e^\frac{\pi i}{2n} \right )^{(2k+1)}}=-\frac{\pi^2}{2n^2}\cdot \csc(\frac{\pi}{2n}) \cdot \cot(\frac{\pi}{2n}\rightarrow (**)) \color{#3D99F6} \text{ see note} }$

$\text{from (*) and (**) } \\ \Rightarrow 2\int _{0}^{\infty}{\frac{\ln(x)}{1+x^{2n}}dx}=-\frac{\pi^2}{2n^2}\cdot \csc(\frac{\pi}{2n}) \cdot \cot(\frac{\pi}{2n}) \\ \Rightarrow \int _{0}^{\infty}{\frac{\ln(x)}{1+x^{2n}}dx}=-\frac{\pi^2}{4n^2}\cdot \csc(\frac{\pi}{2n}) \cdot \cot(\frac{\pi}{2n}) \\ (f\circ g)(n)=-\frac{\pi^2}{4n^2}\cdot \csc(\frac{\pi}{2n}) \cdot \cot(\frac{\pi}{2n}) \\ g(n) \text{is bijective function} \\ g^{-1}=\frac{n}{2} \\ \Rightarrow f(n)=\left [ (f\circ g)\circ g^{-1} \right ](n)=-\frac{\pi^2}{n^2}\cdot \csc(\frac{\pi}{n}) \cdot \cot(\frac{\pi}{n}) \\ \text{now } \frac{f(\frac{3}{2})}{f(3)}=\frac{-\frac{4\pi^2}{9}\cdot \csc(\frac{2\pi}{3}) \cdot \cot(\frac{2\pi}{3})}{-\frac{\pi^2}{9}\cdot \csc(\frac{\pi}{3}) \cdot \cot(\frac{\pi}{3})}=-4$

$\color{#3D99F6} { note \\ \begin{aligned} &\sum_{k=0}^{n-1}{\left ( e^{i\theta} \right )^{2k}} & = & \sum_{k=0}^{n-1}{\left ( e^{2i\theta} \right )^{k}} & = & \frac{\left ( e^{2i\theta} \right)^n -1}{e^{2i\theta}-1} & = & \frac{-2}{e^{2i\theta}-1} & // \text{geometric progression} & // \left ( e^{2i\theta} \right)^n=e^{i(2k+1)\pi} =-1 \\ &\sum_{k=0}^{n-1}{\left ( e^{i\theta} \right )^{2k+1}} & = & \frac{-2 \cdot e^{i\theta}}{e^{2i\theta}-1} & = & \frac{-2}{2i\left ( \frac{e^{i\theta} - e^{-i\theta}}{2i} \right )} & = & i \cdot \csc(\theta) & // \frac{e^{i\theta} - e^{-i\theta}}{2i}=\sin(\theta) \\ &\sum_{k=0}^{n-1}{i(2k+1)\left( e^{i\theta} \right )^{2k+1}} & = & i \cdot \csc(\theta) \cdot \cot(\theta) & & & & & // \text{differentiate both sides} \\ &\sum_{k=0}^{n-1}{(2k+1)\left( e^{i\theta} \right )^{2k+1}} & = & \csc(\theta) \cdot \cot(\theta) \end{aligned} }$