Chứng minh rằng:  $ – arctanx + arctan\frac{{1 + x}}{{1 – x}} = \frac{\pi }{4}$ với $\forall x \in ( – \infty ;1)$

Xét hàm $f(x) =  – {\rm{ar}}ctan + {\rm{ar}}ctan\frac{{1 + x}}{{1 – x}}$ trong  khoảng $\left( { – \infty ;1} \right)$, ta có:
$\begin{array}{l}
f'(x) = – \frac{1}{{1 + {x^2}}} + \frac{1}{{1 + {{\left( {\frac{{1 + x}}{{1 – x}}} \right)}^2}}}.\left( {\frac{{1 + x}}{{1 – x}}} \right)
 = – \frac{1}{{1 + {x^2}}} + \frac{{{{(1 – x)}^2}}}{{2{{(1 + x)}^2}}}.\frac{2}{{{{\left( {1 – x} \right)}^2}}} = 0
\end{array}$
$ \Rightarrow f(x) = C,    \forall x \in \left( { – \infty ;1} \right)$.

Mặt khác,     $f(0) = – {\rm{ar}}ctan0 + {\rm{ar}}ctan1 = \pi /4$
Vậy $f(x) = \pi /4,\forall x \in \left( { – \infty ;1} \right)$