What is the solution for (dy)/(dx)=(x+1)/((x^2+2x-3)^2) ?

The solution for (dy)/(dx)=(x+1)/((x^2+2x-3)^2) is y=-1/(2(x^2+2x-3))+c_{1}

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(dy)/(dx)=(x+1)/((x^2+2x-3)^2)

\frac{d y}{d x} = \frac{x + 1}{( x ^{2} + 2 x - 3 ) ^{2}}

y = - \frac{1}{2 ( x ^{2} + 2 x - 3 )} + c_{1}

Solution steps

\frac{d y}{d x} = \frac{x + 1}{( x ^{2} + 2 x - 3 ) ^{2}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{x + 1}{( x ^{2} + 2 x - 3 ) ^{2}}

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

then

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

y = \int \frac{x + 1}{( x ^{2} + 2 x - 3 ) ^{2}} d x

\int \frac{x + 1}{( x ^{2} + 2 x - 3 ) ^{2}} d x = - \frac{1}{2 ( x ^{2} + 2 x - 3 )} + c_{1}

y = - \frac{1}{2 ( x ^{2} + 2 x - 3 )} + c_{1}

\frac{d}{d t} (9 cos (t) - cos (9 t))

\int e^{\frac{4 x}{3}} (\frac{4}{3}) d x

- \frac{3 y}{3 x + 2 y}

\int 41 tan^{5} (x) sec^{4} (x) d x

x \to 6 lim (\frac{x - 6}{x ^{2} - x})

What is the solution for (dy)/(dx)=(x+1)/((x^2+2x-3)^2) ?
The solution for (dy)/(dx)=(x+1)/((x^2+2x-3)^2) is y=-1/(2(x^2+2x-3))+c_{1}