$\int{\frac{y}{\sqrt{1 + {y}^{2}}}}dy = -\int{\frac{x}{\sqrt{1 + {x}^{2}}}}dx$

asked by guest
on Nov 09, 2024 at 10:32 pm



You asked:

Investigate the equation: \(\int \frac{y}{\sqrt{1 + {y}^{2}}} \cdot d y = -\int \frac{x}{\sqrt{1 + {x}^{2}}} \cdot d x\).

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