Solve the IVP (x+y)^2 dx+(2xy+x^2-1) dy = 0; y(1)=1?

I don't know what to do with this one! It isn't separable and I have already tried solving the DE through exact means (I found that it was exact but was unable to get the correct answers.

The answers is supposed to be (1/3)x^3 + x^2y + xy^2 - y = 4/3

Please help!

P(x,y) = (x+y)^2
Q(2xy + x^2 - 1)

dP/dy = 2x + 2y

dQ/dx = 2y + 2x

this is nice as their equal

choose to integrate (x+y)^2 due to according integrate to respect of x

int (x+y)^2 dx = 1/3*(x + y)^3 + g(y)

now turn it back as (x + y)^2 + dg(y)/dy and set equal to Q(x,y)

(x + y)^2 + dg(y)/y = 2xy + x^2 - 1

solve for dg(y)/dy

dg(y)/dy = 2xy + x^2 - 1 - (x + y)^2

dg(y)/dy = 2xy + x^2 - 1 - x^2 - 2xy - y^2

dg(y)/dy = -y^2 - 1

dg(y) = -y^2 - 1 dy

g(y) = -y^3/3 - y

put your g(y) into 1/3*(x + y)^3 + g(y)

1/3*(x + y)^3 - y^3/3 - y = C

now plug (1,1)

1/3*(1 + 1)^3 - 1^3/3 - 1 = C

C = 4/3

final answer is the top