Cho \(\int\limits_0^{2017} {f\left( x \right){\rm{d}}x}  = 2\).

* Hướng dẫn giải

Đặt \(t = \ln \left( {{x^2} + 1} \right),\) suy ra \({\rm{d}}t = \frac{{2x{\rm{d}}x}}{{{x^2} + 1}} \to \frac{{x{\rm{d}}x}}{{{x^2} + 1}} = \frac{{{\rm{d}}t}}{2}.\)

Đổi cận: \(\left\{ \begin{array}{l}
x = 0 \to t = 0\\
x = \sqrt {{e^{2017}} - 1}  \to t = 2017
\end{array} \right..\)

Khi đó \(I = \frac{1}{2}\int\limits_0^{2017} {f\left( t \right){\rm{d}}t}  = \frac{1}{2}\int\limits_0^{2017} {f\left( x \right){\rm{d}}x}  = \frac{1}{2}.2 = 1.\)