What is the solution for y^'=(1+y^2)/(1+x^2) ?

The solution for y^'=(1+y^2)/(1+x^2) is y=tan(arctan(x)+c_{1})

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y^'=(1+y^2)/(1+x^2)

y^{^{'}} = \frac{1 + y ^{2}}{1 + x ^{2}}

y = tan (arctan (x) + c_{1})

Solution steps

y^{'} (x) = \frac{1 + y ^{2}}{1 + x ^{2}}

Solve separable ODE:

y = tan (arctan (x) + c_{1})

y = tan (arctan (x) + c_{1})

\int \frac{1}{s ^{2}} d s

x \to - 1 lim (\frac{x - 1}{x ^{2} - 1})

\int_{1}^{2.5} e^{- x} d x

\frac{d}{d x} (\frac{x ^{2} - 2 x + 1}{x + 1})

\int_{- \infty}^{- 3} x e^{x} d x

What is the solution for y^'=(1+y^2)/(1+x^2) ?
The solution for y^'=(1+y^2)/(1+x^2) is y=tan(arctan(x)+c_{1})