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