My online class provided this solution but i have questions about some of the steps. My online professor has been no help. Any input is appreciated.

Question Solve the initial value problem. $\displaystyle \frac{dy}{dx}=\frac{y-1}{x+3}$

Step 1 Separate and integrate $\displaystyle \int\frac{dy}{y-1}=\int\frac{1}{x+3}dx$

Step 2 Integration result $\displaystyle ln(y-1)=ln(x+3)+lnC$ So where did the$\displaystyle lnC$come from. I know where the$\displaystyle C$came from but why are they multiplying it by$\displaystyle ln$

Step 3 Raise all to the e $\displaystyle e^{ln(y-1)}=e^{ln(x+3)}+e^{lnC}$

Step 4 Result is $\displaystyle (y-1)=C(x+3)$ Why are they multiplying the right side by $\displaystyle C$ shouldnt it be like this $\displaystyle (y-1)=(x+3)+C$

Step 5 $\displaystyle (y-1)=-C(x+3)$ why it the right side now negative?

Step 6 $\displaystyle y=1-C(x+3)$ Solution given to us by the professor.

I cant figure out how they arrived to some of those steps maybe someone here can. Thanks.