What is the solution for y^'=(y^2+1)/(x^2+1),y(0)=1 ?

The solution for y^'=(y^2+1)/(x^2+1),y(0)=1 is y=tan(arctan(x)+pi/4)

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y^'=(y^2+1)/(x^2+1),y(0)=1

y^{^{'}} = \frac{y ^{2} + 1}{x ^{2} + 1}, y (0) = 1

y = tan (arctan (x) + \frac{π}{4})

Solution steps

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

Solve separable ODE:

y = tan (arctan (x) + \frac{π}{4})

y = tan (arctan (x) + \frac{π}{4})

\frac{d y}{d t} = a y

x \to 0 lim (\frac{tan ( 12 x )}{sin ( 3 x )})

n = 1 \sum \infty \frac{2}{n ^{2} + 1}

\int e^{c o s (38 t)} sin (38 t) d t

\frac{d}{d x} (x^{2} + 8 x + 7)

What is the solution for y^'=(y^2+1)/(x^2+1),y(0)=1 ?
The solution for y^'=(y^2+1)/(x^2+1),y(0)=1 is y=tan(arctan(x)+pi/4)