Calculus of a Single Variable, Chapter 6, 6.2, Section 6.2, Problem 3

Recall that in solving simple first order "ordinary differential equation" (ODE), we may apply variable separable differential equation wherein:
N(y)y'=M(x)
N(y)(dy)/(dx)=M(x)
N(y) dy=M(x) dx
Before we can work on the direct integration: int N(y) dy= int M(x) dx to solve for the general solution of a differential equation.
For the given first order ODE: (dy)/(dx)=y+3 can be rearrange by cross-multiplication into:
(dy)/(y+3)=dx
Apply direct integration on both sides: int(dy)/(y+3)=int dx
For the left side, we consider u-substitution by letting:
u= y+3 then du = dy

The integral becomes:
int(dy)/(y+3)=int(du)/(u)
Applying basic integration formula for logarithm:
int(du)/(u)= ln|u|
Plug-in u = y+3 on ln|u | , we get:
int(dy)/(y+3)=ln|y+3|
For the right side, we apply the basic integration: int dx= x+C

Combing the results from both sides, we get the general solution of the differential equation as:
ln|y+3|= x+C
or
y =e^(x+C)-3
y =Ce^x-3