Integrate $\int_{-\infty}^{\infty} \frac{e^{2020x}-e^{x}}{x\left(e^{2020x}+1\right)\left(e^x+1\right)} \mathop{dx}$

Instead of trying to use partial fraction decomposition, how about try to introduce a parameter, $a$, inside the integral? Sometimes when there are integrals with strange numbers, like $2020$ in this case, the integral can be generalized. $$I(a)=\int_{-\infty}^{\infty} \frac{e^{ax}-e^x}{x\left(e^{ax}+1\right) \left(e^x+1\right)} \; \mathrm{d}x$$ Now, factor out the terms independent of $a$, and differentiate both sides with respect to $a$: \begin{align*} I'(a)&=\int_{-\infty}^{\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} \; \mathrm{d}x \\ &=\int_{-\infty}^{\infty} \frac{e^{ax}}{{\left(e^{ax}+1\right)}^2} \; \mathrm{d}x \\ &=-\frac{1}{a\left(e^{ax}+1\right)} \bigg \rvert_{-\infty}^{\infty} \\ &=\frac{1}{a}\\ \end{align*} Integrate both sides with respect to $a$: $$I(a)=\ln{a}+C$$ Notice that $I(1)=0$. $$0=\ln{1}+C \implies C=0$$ Therefore, the integral you posted evaluates to: $$I(2020)=\boxed{\ln{2020}}$$

Let $f(x) =\frac1{e^x+1}$. Then

\begin{align} &\int_{-\infty}^{\infty} \frac{e^{2020x}-e^{x}}{x\left(e^{2020x}+1\right)\left(e^x+1\right)} \mathop{dx}\\ = & \>2\int_{0}^{\infty} \frac{e^{2020x}-e^{x}} {x\left(e^{2020x}+1\right)\left(e^x+1\right)} \mathop{dx} \\ = & \>2\int_{0}^{\infty} \frac{f(x)-f(2020x)}x dx \\ = & \>2[f(0)-f(\infty)] \ln \frac{2020}1\\ =&\> \ln 2020 \end{align}

where the Frullani integral is applied.