What is the solution for (dy)/(dx)=(2x)/(x^2-9) ?

The solution for (dy)/(dx)=(2x)/(x^2-9) is y=ln(x^2-9)+c_{1}

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(dy)/(dx)=(2x)/(x^2-9)

\frac{d y}{d x} = \frac{2 x}{x ^{2} - 9}

y = ln (x^{2} - 9) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{2 x}{x ^{2} - 9}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{2 x}{x ^{2} - 9}

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

then

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

y = \int \frac{2 x}{x ^{2} - 9} d x

\int \frac{2 x}{x ^{2} - 9} d x = ln (x^{2} - 9) + c_{1}

y = ln (x^{2} - 9) + c_{1}

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What is the solution for (dy)/(dx)=(2x)/(x^2-9) ?
The solution for (dy)/(dx)=(2x)/(x^2-9) is y=ln(x^2-9)+c_{1}