# Find the integral of 1/x

We have to find the integral of 1/x

### Solution

From differential calculus, by the fundamental theorem of Calculus, we learned that the derivative of ln(x) is 1/x.

That is $\frac{d}{dx}(\ln x)= \frac{1}{x}$ for x > 0

From chain rule we also know that

$\frac{d}{dx}(\ln (-x))= \frac{1}{x} foe x< 0$

The integral (or antiderivative) of 1/x should be a function whose derivative is 1/x.

As we saw above this is ln(x). However, if x is negative then ln(x) is undefined.

On an interval that excludes 0, the antiderivative of 1/x is ln x if interval lies in positive integers and ln (-x) if intervals lie in negative numbers.

Hence ln |x| covers both negative and positive integers.