$\underset{0}{\overset{1}{\int }}\frac{{\mathrm{x}}^4{(1-2\mathrm{x}+{\mathrm{x}}^2)}^2}{1+{\mathrm{x}}^2}\mathrm{dx}$

N^r = x⁴{(1 + x²)² + 4x² – 4x(1 + x²)}

 $\underset{0}{\overset{1}{\int }}\frac{{\mathrm{x}}^4\left\{{(1+{\mathrm{x}}^2)}^2+4{\mathrm{x}}^2-4\mathrm{x}(1+{\mathrm{x}}^2)\right\}}{1+{\mathrm{x}}^2}\mathrm{dx}=\underset{0}{\overset{1}{\int }}\left({\mathrm{x}}^4(1+{\mathrm{x}}^2)+\frac{4{\mathrm{x}}^6}{1+{\mathrm{x}}^2}-4{\mathrm{x}}^5\right)\mathrm{dx}$

 $=\underset{0}{\overset{1}{\int }}({\mathrm{x}}^4+{\mathrm{x}}^6+\frac{4({\mathrm{x}}^6+1-1)}{1+{\mathrm{x}}^2}-4{\mathrm{x}}^5)\mathrm{dx}=\underset{0}{\overset{1}{\int }}\left({\mathrm{x}}^4+{\mathrm{x}}^6+4({\mathrm{x}}^4-{\mathrm{x}}^2+1)-\frac{4}{1+{\mathrm{x}}^2}-4{\mathrm{x}}^5\right)\mathrm{dx}$

 $=\underset{0}{\overset{1}{\int }}(5{\mathrm{x}}^4+{\mathrm{x}}^6-4{\mathrm{x}}^5-4{\mathrm{x}}^2+4-\frac{4}{1+{\mathrm{x}}^2})\mathrm{dx}={({\mathrm{x}}^5+\frac{{\mathrm{x}}^7}{7}-\frac{4{\mathrm{x}}^6}{6}-\frac{4{\mathrm{x}}^3}{3}+4\mathrm{x}-4{\tan }^{-1}\mathrm{x})}_0^1$

 $1+\frac{1}{7}-\frac{2}{3}-\frac{4}{3}+4-4\mathrm{·}\frac{\mathrm{\pi }}{4}=5+\frac{1}{7}-\frac{6}{3}-\mathrm{\pi }=\frac{22}{7}-\mathrm{\pi }$