Cho f(x) là hàm số chẵn liên tục trong đoạn [-1;1] và tích phân từ -1 đến 1 f(x)dx=2 .

$I=\underset{-1}{\overset{1}{\int }}\frac{f\left(x\right)+2020}{1+e^x}d\text{x}=\underset{-1}{\overset{0}{\int }}\frac{f\left(x\right)+2020}{1+e^x}d\text{x}+\underset{0}{\overset{1}{\int }}\frac{f\left(x\right)+2020}{1+e^x}d\text{x}=I_1+I_2$

Xét $I_1=\underset{-1}{\overset{0}{\int }}\frac{f\left(x\right)+2020}{1+e^x}d\text{x}$. Đặt $x=-t\Rightarrow d\text{x}=-dt$, đổi cận $x=0\Rightarrow t=0,x=-1\Rightarrow t=1$.

$I_1=\underset{1}{\overset{0}{\int }}\frac{f\left(-t\right)+2020}{1+e^{-t}}\left(-dt\right)=\underset{0}{\overset{1}{\int }}\frac{e^t\left[f\left(t\right)+2020\right]}{1+e^t}dt$

Ta có $\underset{0}{\overset{1}{\int }}\frac{e^t\left[f\left(t\right)+2020\right]}{1+e^t}dt=\underset{0}{\overset{1}{\int }}\frac{e^x\left[f\left(x\right)+2020\right]}{1+e^x}d\text{x}$.

Suy ra $I=\underset{-1}{\overset{1}{\int }}\frac{f\left(x\right)+2020}{1+e^x}d\text{x}=\underset{0}{\overset{1}{\int }}\frac{e^x\left[f\left(x\right)+2020\right]}{1+e^x}d\text{x}+\underset{0}{\overset{1}{\int }}\frac{f\left(x\right)+2020}{1+e^x}d\text{x}$

$=\underset{0}{\overset{1}{\int }}\frac{\left(1+e^x\right)\left[f\left(x\right)+2020\right]}{1+e^x}d\text{x}=\underset{0}{\overset{1}{\int }}\left[f\left(x\right)+2020\right]d\text{x}=\frac{1}{2}\underset{-1}{\overset{1}{\int }}\left[f\left(x\right)+2020\right]d\text{x}=2021$.