$f(x)$ là hàm lẻ liên tục trên $[-1;1]$
$\Rightarrow f(-x)=-f(x)\\ g(x)=\dfrac{f(x)}{x^2+1}\\ g(-x)=\dfrac{f(-x)}{x^2+1}=-\dfrac{f(x)}{x^2+1}=-g(x)$
$\Rightarrow g(x)$ cũng là hàm lẻ liên tục trên $[-1;1]$
$\Rightarrow \displaystyle\int\limits^1_{-1} g(x) \, dx=0\\ \displaystyle\int\limits^1_{-1} \dfrac{f(x)+1}{x^2+1} \, dx\\ I=\displaystyle\int\limits^1_{-1} \dfrac{f(x)}{x^2+1} \, dx+\displaystyle\int\limits^1_{-1} \dfrac{1}{x^2+1} \, dx\\ =\displaystyle\int\limits^1_{-1} g(x) \, dx+\displaystyle\int\limits^1_{-1} \dfrac{1}{x^2+1} \, dx\\ =\displaystyle\int\limits^1_{-1} \dfrac{1}{x^2+1} \, dx\\ x=\tan t \\ dx=(\tan^2t+1)dt\\ \text{Đổi cận}\\ \begin{array}{|c|ccc|} \hline x&-1&&1\\\hline t&-\dfrac{\pi}{4} && \dfrac{\pi}{4}\\\hline\end{array} \\ I=\displaystyle\int\limits^{\tfrac{\pi}{4}}_{-\tfrac{\pi}{4}} \dfrac{1}{\tan^2t+1+1} \, (\tan^2t+1)dt\\ =\displaystyle\int\limits^{\tfrac{\pi}{4}}_{-\tfrac{\pi}{4}} \, dt\\ =t \Bigg\vert^{\tfrac{\pi}{4}}_{-\tfrac{\pi}{4}}\\ =\dfrac{\pi}{2}\\ \Rightarrow C$
