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

Jul 11, 2016

$f \left(x\right) = \ln \left(\frac{x + 3}{5}\right) + 1$

#### Explanation:

We know that $\int \frac{1}{x} \mathrm{dx} = \ln x + C$, so:
$\int \frac{1}{x + 3} \mathrm{dx} = \ln \left(x + 3\right) + C$

Therefore $f \left(x\right) = \ln \left(x + 3\right) + C$. We are given the initial condition $f \left(2\right) = 1$. Making necessary substitutions, we have:
$f \left(x\right) = \ln \left(x + 3\right) + C$
$\to 1 = \ln \left(\left(2\right) + 3\right) + C$
$\to 1 - \ln 5 = C$

We can now rewrite $f \left(x\right)$ as $f \left(x\right) = \ln \left(x + 3\right) + 1 - \ln 5$, and that is our final answer. If you want to, you can use the following natural log property to simplify:
$\ln a - \ln b = \ln \left(\frac{a}{b}\right)$

Applying this to $\ln \left(x + 3\right) - \ln 5$, we obtain $\ln \left(\frac{x + 3}{5}\right)$, so we can further express our answer as $f \left(x\right) = \ln \left(\frac{x + 3}{5}\right) + 1$.