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

1 Answer
Jul 11, 2016

Answer:

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

Explanation:

We know that #int1/xdx=lnx+C#, so:
#int1/(x+3)dx=ln(x+3)+C#

Therefore #f(x)=ln(x+3)+C#. We are given the initial condition #f(2)=1#. Making necessary substitutions, we have:
#f(x)=ln(x+3)+C#
#->1=ln((2)+3)+C#
#->1-ln5=C#

We can now rewrite #f(x)# as #f(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)#

Applying this to #ln(x+3)-ln5#, we obtain #ln((x+3)/5)#, so we can further express our answer as #f(x)=ln((x+3)/5)+1#.