Question

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

Answer 1

`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`.