Cho \(f\left( x \right)\) là hàm số chẵn, liên tục trên đoạn \(\left[ -1;1 \right]\) và \(\int\limits_{-1}^{1}{f\left( x \right)dx}=4\). Kết quả \(I=\int\limits_{-1}^{1}{\frac{f\left( x \right)}{1+{{e}^{x}}}dx}\) bằng:

Đặt \(t=-x\Rightarrow dt=-dx\).

Đổi cận \(\left\{ \begin{align}

& x=1\Rightarrow t=-1 \\

& x=-1\Rightarrow t=1 \\

\end{align} \right.\), khi đó:

\(I=\int\limits_{-1}^{1}{\frac{f\left( x \right)}{1+{{e}^{x}}}dx}=-\int\limits_{1}^{-1}{\frac{f\left( -t \right)dt}{1+{{e}^{-t}}}}=\int\limits_{-1}^{1}{\frac{f\left( -x \right)dx}{1+\frac{1}{{{e}^{x}}}}}=\int\limits_{-1}^{1}{\frac{{{e}^{x}}f\left( -x \right)dx}{1+{{e}^{x}}}}\)

Do \(f\left( x \right)\) là hàm số chẵn nên \(f\left( x \right)=f\left( -x \right)\text{ }\forall x\in \left[ -1;1 \right]\Rightarrow I=\int\limits_{-1}^{1}{\frac{{{e}^{x}}f\left( x \right)}{1+{{e}^{x}}}dx}\)

\(\Rightarrow I+I=\int\limits_{-1}^{1}{\frac{f\left( x \right)}{1+{{e}^{x}}}dx}+\int\limits_{-1}^{1}{\frac{{{e}^{x}}f\left( x \right)}{1+{{e}^{x}}}}dx=\int\limits_{-1}^{1}{\frac{\left( {{e}^{x}}+1 \right)f\left( x \right)dx}{1+{{e}^{x}}}}=\int\limits_{-1}^{1}{f\left( x \right)dx}=4\Rightarrow I=2\).