[tex]\text{Formula lui King:}\\ \displaystyle \boxed{\int_{a}^b f(x)\, dx = \int_{a}^b f(a+b-x)\, dx}[/tex]
[tex]\displaystyle I = \int_{-1}^{1} \dfrac{x^2}{1+e^x}\, dx \\ \\ I =\int_{-1}^1 \dfrac{(-1+1-x)^2}{1+e^{-1+1-x}}\,dx = \int_{-1}^1 \dfrac{x^2}{1+e^{-x}}\, dx =\int_{-1}^1 \dfrac{x^2}{1+\dfrac{1}{e^x}}\, dx = \\ \\ = \int_{-1}^1 \dfrac{x^2}{\dfrac{e^x+1}{e^x}}\, dx = \int_{-1}^1 \dfrac{e^xx^2}{1+e^x}\,dx\\ \\ \\ I+I = \int_{-1}^{1} \dfrac{x^2}{1+e^x}\, dx+\int_{-1}^1 \dfrac{e^xx^2}{1+e^{x}}\, dx = \int_{-1}^1 \dfrac{x^2+e^xx^2}{1+e^x}\, dx =[/tex]
[tex]\displaystyle = \int_{-1}^1 \dfrac{x^2(1+e^x)}{1+e^x}\, dx = \int_{-1}^1{x^2}\, dx = \dfrac{x^3}{3}\Big|_{-1}^1 = \dfrac{1}{3}-\dfrac{-1}{3} = \dfrac{2}{3} \\ \\ 2I = \dfrac{2}{3} \Rightarrow \boxed{I = \dfrac{1}{3}}[/tex]
