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

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

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(1-x^2)y^'=1

(1 - x^{2}) y^{^{'}} = 1

y = \frac{ln ( x + 1 )}{2} - \frac{ln ( x - 1 )}{2} + c_{1}

Solution steps

(1 - x^{2}) y^{'} (x) = 1

Isolate

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{1}{1 - x ^{2}} d x

\int \frac{1}{1 - x ^{2}} d x = \frac{ln ( x + 1 )}{2} - \frac{ln ( x - 1 )}{2} + c_{1}

y = \frac{ln ( x + 1 )}{2} - \frac{ln ( x - 1 )}{2} + c_{1}

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