We know that int1/xdx=lnx+C∫1xdx=lnx+C, so:
int1/(x+3)dx=ln(x+3)+C∫1x+3dx=ln(x+3)+C
Therefore f(x)=ln(x+3)+Cf(x)=ln(x+3)+C. We are given the initial condition f(2)=1f(2)=1. Making necessary substitutions, we have:
f(x)=ln(x+3)+Cf(x)=ln(x+3)+C
->1=ln((2)+3)+C→1=ln((2)+3)+C
->1-ln5=C→1−ln5=C
We can now rewrite f(x)f(x) as f(x)=ln(x+3)+1-ln5f(x)=ln(x+3)+1−ln5, and that is our final answer. If you want to, you can use the following natural log property to simplify:
lna-lnb=ln(a/b)lna−lnb=ln(ab)
Applying this to ln(x+3)-ln5ln(x+3)−ln5, we obtain ln((x+3)/5)ln(x+35), so we can further express our answer as f(x)=ln((x+3)/5)+1f(x)=ln(x+35)+1.