What is f(x) = int 1/(x+3) f(x)=1x+3 if f(2)=1 f(2)=1?

1 Answer
Jul 11, 2016

f(x)=ln((x+3)/5)+1f(x)=ln(x+35)+1

Explanation:

We know that int1/xdx=lnx+C1xdx=lnx+C, so:
int1/(x+3)dx=ln(x+3)+C1x+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)+C1=ln((2)+3)+C
->1-ln5=C1ln5=C

We can now rewrite f(x)f(x) as f(x)=ln(x+3)+1-ln5f(x)=ln(x+3)+1ln5, 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)lnalnb=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.