[tex]\displaystyle\int\frac{x^7}{1+x^{16}}\,\mathrm dx[/tex]
Notice that [tex]x^{16}=(x^8)^2[/tex], so let's make the substitution [tex]y=x^8[/tex], so that [tex]\mathrm dy=8x^7\,\mathrm dx[/tex]. Then the above can be written as
[tex]\displaystyle\frac18\int\frac{\mathrm dy}{1+y^2}=\frac18\tan^{-1}y+C=\frac18\tan^{-1}x^8+C[/tex]
