## FANDOM

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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

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