Question

Find the integral of $\frac{1}{x}$.

Answer 1

Find the integral

By the fundamental theorem of Calculus, we know that the derivative of $\ln \left(x\right)$is $\frac{1}{x}$.

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

From chain rule we also know that,

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

The integral of $\frac{1}{x}$ should be a function whose derivative is $\frac{1}{x}$.

So we can say,

However, if $x$ is negative then $\ln \left(x\right)$is undefined.

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

Hence, $\ln \left|x\right|$covers both negative and positive integers.

So, $\int \frac{1}{x}dx=\ln \left|x\right|+c$ where, $c$ is the integration constant.

Hence, The integral of $\frac{1}{x}$ is $\ln \left|x\right|+c$.