Solve `(x^(2) + 2xy + y^(2) + 1)(dy)/(dx) = 2(x+y)`

Explanation

Observe that `x^(2) + 2xy + y^(2) = (x+y)^(2)`
The equation becomes,
`{(x+y)^(2)+1}(dy)/(dx) = 2(x+y)`
Set x+ y = t, so that `1+(dy)/(dx) = (dt)/(dx)` (on differentiation w.r.t. x)
Now, `(t^(2)+1)((dt)/(dx) - 1) = 2t`
`rArr (dt)/(dx) = (2t)/(t^(2) + 1) +1 = (t^(2) + 2t +1)/(t^(2)+1) = (t^(2) + 2t + 1)/(t^(2) + 1) = ((t + 1)^(2))/(t^(2)+1)`
`rArr (t^(2) + 1)/((t + 1)^(2))dt = dx`
`rArr ((t+1)^(2)-2(t+1)+2)/((t+1)^(2)) dt = dx`
`rArr dt - (2)/(t+1)dt + 2(dt)/((1+t)^(2)) = dx`
Integrating, we obtain,
`t-2ln |t+1| - (2)/(t-1) = x + k`
`rArr x+y - 2ln|x+y+1| - (1)/(x+y+1) = x+k`
`:. y- 2ln|x+y+1| - (2)/(x+y+1) = k`, k being the constant of integration