A student asks, I need to learn about the integral of 1/x.

## solutionEdit

Let's consider the indefinite integral,

- $ \int \! {\frac {1} {x}} \,dx \, $

The solution is $ \log(x)+C $

Now, let's consider the definite integral,

- $ \int_a^b \! {\frac {1} {x}} \,dx \, $

Evaluating the indefinite integral at its limits, we get

- $ \log(x)+C \: \bigg{|}_a^b = \log(b) - \log(a) $