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

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

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1/y dy=(x+1)/(1-x^2)dx

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

y = \frac{c _{1}}{x - 1}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{x - 1}

y = \frac{c _{1}}{x - 1}

y = 4 x^{4} + \frac{1}{3} x^{3} + x + 1

\int \frac{1}{x ^{2} - 3 x + 2} d x

\int \frac{4 x + 4}{2 x + 1} d x

\frac{d y}{d t} + 2 t (y - 5) = 0

\frac{d}{d x} (2 (ln (x))^{3})

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