Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=\int^\infty_0\frac{\log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where
$$I(a)=\int^\infty_{0}\frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^a\equiv \exp(a(\ln|z|+i\arg z))$ where $\arg z\in[0,2\pi)$.
Firstly, by residue theorem,
$$\oint_C f(z)dz=2\pi i\bigg(\operatorname*{Res}_{z=i}f(z)+\operatorname*{Res}_{z=-i}f(z)\bigg)$$
We have
$$\operatorname*{Res}_{z=i}f(z)=\frac{\exp(a(\ln|i|+i\arg i))}{i+i}=\frac{e^{\pi ia/2}}{2i}$$
$$\operatorname*{Res}_{z=-i}f(z)=\frac{\exp(a(\ln|-i|+i\arg -i))}{-i-i}=-\frac{e^{3\pi ia/2}}{2i}$$
Thus,
$$\oint_C f(z)dz=\pi(e^{\pi ia/2}-e^{3\pi ia/2})$$
On the other hand,
$$\oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$
K_1=\lim_{R\to\infty}\int^{2\pi}_0 f(Re^{it})iRe^{it}dt
=\lim_{R\to\infty}2\pi f(Re^{ic})iRe^{ic}=0 \qquad{c\in[0,2\pi]}$$
$$K_2=\lim_{r\to0^+}\int_{2\pi}^0 f(re^{it})ire^{it}dt
=\lim_{r\to0^+}2\pi f(re^{ic})ire^{ic}=0 \qquad{c\in[0,2\pi]}$$
$$K_3=\int^\infty_0 f(te^{i0})dt=\int^\infty_0\frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=\int_\infty^0 f(te^{i2\pi})dt=-\int^\infty_0\frac{e^{2\pi ia}t^a}{t^2+1}dt=-e^{2\pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)\sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore,
$$I-e^{2\pi ia}I=\pi(e^{\pi ia/2}-e^{3\pi ia/2})$$
$$\implies I=\pi\frac{e^{\pi ia/2}-e^{3\pi ia/2}}{1-e^{2\pi ia}}
=\pi\frac{e^{-\pi ia/2}-e^{\pi ia/2}}{e^{-\pi i a}-e^{\pi ia}}
=\pi\frac{\sin(\pi a/2)}{\sin(\pi a)}
=\frac{\pi}2\sec\left(\frac{\pi a}2\right)
$$
Let $T=\tan(\pi a/2)$, $S=\sec(\pi a/2)$.
$$I^{(4)}(a)=\frac{\pi}2\frac{\pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence,
$$J=I^{(4)}(0)=\frac{\pi}2\frac{\pi^4(0+0+5\cdot 1)}{16}=\color{red}{\frac{5\pi^5}{32}}$$
The tedious differentiation is done by calculator. :)
Best Answer
Let $f(x) = \int_{x}^{\infty} \frac{\mathrm{d}t}{1+t^3}$. Then $f$ is integrable on $[0, \infty)$. Indeed, $f$ is bounded by $f(0)$ and $f(x) \asymp 1/x^2$ as $x\to\infty$. Now by integration by parts,
\begin{align*} \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x &= -\int_{0}^{\infty} n^2(\cos(x/n^2)-1)f'(x) \,\mathrm{d}x \\ &= \underbrace{\left[ -n^2(\cos(x/n^2)-1)f(x) \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \sin(x/n^2)f(x) \, \mathrm{d}x. \end{align*}
Now by the dominated convergence theorem,
$$ \lim_{n\to\infty} \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x = - \int_{0}^{\infty} \lim_{n\to\infty} \sin(x/n^2)f(x) \, \mathrm{d}x = 0. $$
Addendum. A more detailed analysis, with a bit of help from Mathematica 11, shows that
$$ \int_{0}^{\infty} \frac{n^2(\cos(x/n^2)-1)}{1+x^3}\,\mathrm{d}x = -\frac{1}{n^2}\left(\log n + \frac{3}{4} - \frac{\gamma}{2} + o(1) \right) $$
as $n\to\infty$, where $\gamma$ is the Euler-Mascheroni constant.