$\int_ {0}^{\infty} \frac{(e^{3x}-e^x)dx}{x(e^x+1)(e^{3x}+1)}$

You were on the right track. The integral separates out into

$$\int_0^\infty \frac{dx}{x}\left(\frac{1}{e^x+1}-\frac{1}{e^{3x}+1}\right) = \int_0^ \infty\int_1^3 \frac{e^{xy}}{(e^{xy}+1)^2}\:dy\:dx$$

Then we can swap the order of integration to get

$$\int_1^3 \frac{dy}{y}\frac{-1}{e^{xy}+1}\Biggr|_0^\infty = \int_1^3\frac{dy}{2y} = \frac{\log 3}{2}$$

We can write $$I=\int_{0}^{\infty}\frac{\phi(x)-\phi(3x)}{x} dx, ~~~\phi(x)=\frac{1}{1+e^x}$$ Next we use the ideas of Frullani's Integration formula:

https://en.wikipedia.org/wiki/Frullani_integral

Since $\int^{\infty}_0 \phi(x) dx$ is convergent at $\infty$, then $$\int_{0}^{\infty} \frac{\phi(ax)-\phi(bx)}{x} dx=\phi_0 \ln \frac{b}{a},~~ \phi_0=\lim_{x\to 0} \phi(x)=\frac{1}{2}.$$ So $$I=\frac{\ln 3}{2}$$

I will present a simple solution using Feynman's method, as you asked for. Consider the following parameterized integral: $$I(a)=\int_{0}^{\infty} \frac{e^{ax}-e^x}{x\left(e^{ax}+1\right) \left(e^x+1\right)} \; dx$$ Now, differentiate both sides with respect to $a$ (factor out the terms independent of $a$ for simplicity): $$I'(a)=\int_{0}^{\infty} \frac{1}{x \left(e^x+1\right)} \cdot \frac{x e^{ax}\left(e^{ax}+1\right)-xe^{ax}\left(e^{ax}-e^x\right)}{{\left(e^{ax}+1\right)}^2} \; dx$$ Simplifying this yields an elegant integral: $$I'(a)=\int_{0}^{\infty} \frac{e^{ax}}{{\left(e^{ax}+1\right)}^2} \; dx$$ $$I'(a)=-\frac{1}{a\left(e^{ax}+1\right)} \bigg \rvert_{0}^{\infty}$$ $$I'(a)=\frac{1}{2a}$$ Integrate both sides with respect to $a$: $$I(a)=\frac{\ln{a}}{2}+C$$ If you plug in $a=1$ into the original, you get that $I(1)=0$. $$0=\frac{\ln{1}}{2}+C \implies C=0$$

Therefore, your integral is: $$\boxed{I=I(3)=\frac{\ln{3}}{2}}$$