Solve: (3xy+y) dx + (x + xy) dy = 0

Please give the steps after integration begins.

there is no need for long integration method just open the brackets

3xydx + ydx + xdy+xydy = 0

ydx+xdy=-(3xydx+xydy)

ydx+xdy/xy = 3dx+dy

now see the lhs this is the differential of logxy (u have to memorise some certain forms like these plus u can check it by differentiating logxy)

now integrating it becomes

logxy = 3x + y + c

can covert to a required form if needed

solution:

=>(3xy+y)dx+(x+xy) dy=0

=>(3x+1) y dx +(1+y) x dy=0

=>(3x+1) y dx =- (1+y) x dy

=>((3x+1)/ x) dx=-((1+y)/y) dy

=>(3+1/x)dx=(-1/y -1)dy

then intigrate by parts ....

=>3x+lnx=-lny-y

answer is   3x+y+lnxy=0.