Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?

$(y+1)dx+(x+1)dy=0$
$M(x, y) =y+1$
$N(x, y) =x+1$
Then, $\frac {\partial M}{\partial y}=1=\frac {\partial N}{\partial x}$
Hence, the differential equation is exact.
Choose, $u(x, y) $ be such that
$u_x =M $ and $u_y =N$
Then, $u(x, y) =\int {M {dx}}=x(y+1)+h(y) $
$u_y (x, y)=x+h'(y) $
$x+h'(y)=x+1 $
$h(y) =y$
Hence, $u(x, y) =x(y+1)+y $
Solution : $u(x, y) =C$
$\implies x(y+1)+y =C$
$\implies x(y+1)+y+1 =C+1$
$ (y+1)(x+1 )=c \space \space  [c=C+1]$
Hence, $(y+1)=\frac{c}{(x+1)}$
Note:
\begin{align} \frac {d}{dx}{u(x, y(x))} &=u_x + u_y \frac {dy}{dx}\\&=M+N \frac {dy}{dx} =0 \end{align}
Hence,$ u(x, y) =C$
