Is the answer to integrate x/x+3dx this:
ln|x+3|+c ????
Just observe carefully.... there is a x term in the numerator. So according to you u = x+3, du = dx.
So $\displaystyle \int \frac{x}{x+3} \, dx = \int \frac{x}{u} \, du$
But hey you have to get rid of that x in the above expression
Thats why write $\displaystyle \frac{x}{x+3} = 1 - \frac{3}{x+3}$. Both the terms now are easily integrable