[tex]t=\mathrm{arctg}x\Rightarrow \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{1}{x^2+1}\Rightarrow I=\int \frac{\mathrm{arctg}^4x}{x^2+1}=\int t^4dt=\frac{t^5}{5}+C=\frac{\mathrm{arctg}^5x}{5}+C\\J=\int\limits_{0}^{\infty }\frac{\mathrm{arctg}^4x}{1+x^2}=\frac{1}{5}\lim\limits_{x\to \infty }\mathrm{arctg}^5x-\frac{1}{5}\mathrm{arctg}^50=\frac{\pi^5}{160}[/tex]