Rozwiązanie:
[tex]$\int {\frac{1}{(16+\arctan^{2} x)(1+x^{2})} } \, dx =\left|\begin{array}{c}u=\arctan x\\du=\frac{1}{1+x^{2}} \ dx \end{array}\right|=\int {\frac{1}{u^{2}+16} } \, du=[/tex]
[tex]$=\int {\frac{1}{u^{2}+4^{2}} } \, du =\frac{1}{4} \arctan \Big(\frac{u}{4} \Big)+C=\frac{1}{4} \arctan \Big(\frac{\arctan x}{4} \Big)+C[/tex]
