The correct option is
A -1
(x2+x+1)dy+(y2+y+1)dx=0
⇒dy(y2+y+1)=−dx(x2+x+1)
⇒∫dy(y2+y+1)=−∫dx(x2+x+1)
⇒∫dy(y+12)2+(√32)2=−∫dx(x+12)2+(√32)2
⇒2√3tan−1⎛⎜
⎜
⎜
⎜⎝y+12√32⎞⎟
⎟
⎟
⎟⎠+2√3tan−1⎛⎜
⎜
⎜
⎜⎝x+12√32⎞⎟
⎟
⎟
⎟⎠=c
For ease of solving let c=2√3tan−1c where c is a constant
⇒tan−1⎛⎜
⎜
⎜
⎜⎝y+12√32⎞⎟
⎟
⎟
⎟⎠+tan−1⎛⎜
⎜
⎜
⎜⎝x+12√32⎞⎟
⎟
⎟
⎟⎠=tan−1c
We know that
tan−1x+tan−1y=tan−1x+y1+xy
On solving for c, we get
c=2√3(x+y+1)−4xy−2y−2x+2(x+y+1)=c√3(1−x−y−2xy)
Given general solution of the equation
(x+y+1)=A(1+Bx+Cy+Dxy)
Comparing, we get
B=−1