How do you solve the following separable differential equation: y'y = y + 1? I just started learning about differential equations and encountered following equation:
$$ y'y = y +1 $$
Wolfram alpha provided the following explanation:
here
But I'm not sure how the integration is performed. What rules are used?
How does $\int{\frac{y'y}{y+1}dx}$ become $-\log(y + 1) + y$?

$$\frac{dy}{dx}y=y+1$$
$$\int\frac{y}{y+1}dy=\int dx$$
$$\int1-\frac{1}{y+1}dy=\int dx$$
$$y-\log(y+1)=x+c$$
