What is the integral of (y/((x+y)^2)-1) ?

The integral of (y/((x+y)^2)-1) is -x-y/(y+x)+C

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integral of (y/((x+y)^2)-1)

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

- x - \frac{y}{y + x} + C

Solution steps

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

Apply the Sum Rule:

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

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

\int 1 d x = x

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

= - x - \frac{y}{y + x}

Add a constant to the solution

= - x - \frac{y}{y + x} + C

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What is the integral of (y/((x+y)^2)-1) ?
The integral of (y/((x+y)^2)-1) is -x-y/(y+x)+C