Methodus inveniendi lineas curvas

AD CURVAS INVENIENDAS ABSOLUTA. 55 est $k(y+x\sqrt{-1})^{2n+1} - k(y-x\sqrt{-1})^{2n+1} = \frac{1}{\sqrt{-1}}(y+x\sqrt{-1})^{2n+1} - \frac{1}{\sqrt{-1}}(y-x\sqrt{-1})^{2n+1} + C$, seu $C = (y+x\sqrt{-1})^{2n+1} (k\sqrt{-1}+1) + (y-x\sqrt{-1})^{2n+1} (1-k\sqrt{-1})$. At est generaliter $(y+x\sqrt{-1})^{2n+1} + (y-x\sqrt{-1})^{2n+1} = 2(yy+xx)^{(2n+1):2} \text{cos. } 2n \text{ A tang. } \frac{x}{y}$, atque $\frac{(y+x\sqrt{-1})^{2n+1} - (y-x\sqrt{-1})^{2n+1}}{\sqrt{-1}} = 2(yy+xx)^{(2n+1):2} \text{sin. } 2n \text{ A tang. } \frac{x}{y}$. Quibus valoribus substitutis, prodibit aequatio integralis ab imaginariis libera haec $2k(yy+xx)^{( [TRUNCATED]