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}

x \to 9 lim (12)

\int x^{2} (1 + x^{3}) d x

\int \frac{4}{3} t^{3} - 4 t^{2} + 3 t + 10 d t

\frac{d}{d t} (t + 1)

x \to 5 + lim (\frac{2}{x ^{2} - 25})

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}