Đặt \(ln\left(x^2+1\right)=t\Rightarrow\dfrac{x}{x^2+1}dx=\dfrac{1}{2}dt\)
\(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=\sqrt{e^{23}-1}\Rightarrow t=23\end{matrix}\right.\)
\(I=\dfrac{1}{2}\int\limits^{23}_0f\left(t\right)dt=\dfrac{1}{2}\int\limits^{23}_0f\left(x\right)dx=1\)