What is the solution for (dy)/(dx)= 4/(x^2-1) ?

The solution for (dy)/(dx)= 4/(x^2-1) is y=-4(1/2 ln(x+1)-1/2 ln(x-1))+c_{1}

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(dy)/(dx)= 4/(x^2-1)

\frac{d y}{d x} = \frac{4}{x ^{2} - 1}

y = - 4 (\frac{1}{2} ln (x + 1) - \frac{1}{2} ln (x - 1)) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{4}{x ^{2} - 1}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{4}{x ^{2} - 1}

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

then

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

y = \int \frac{4}{x ^{2} - 1} d x

\int \frac{4}{x ^{2} - 1} d x = - 4 (\frac{1}{2} ln (x + 1) - \frac{1}{2} ln (x - 1)) + c_{1}

y = - 4 (\frac{1}{2} ln (x + 1) - \frac{1}{2} ln (x - 1)) + c_{1}

x \to 1 + lim (\frac{1 + x}{1 - x})

\frac{d}{d x} (6 e)

\int \frac{- 4 x - 10}{x ^{3}} d x

f (x) = x^{2} - 12, g (x) = x - 6, x = - 2, x = 3

\int 3 x^{7} + 4 x^{5} - 2 x^{3} - x + 1 d x

What is the solution for (dy)/(dx)= 4/(x^2-1) ?
The solution for (dy)/(dx)= 4/(x^2-1) is y=-4(1/2 ln(x+1)-1/2 ln(x-1))+c_{1}