# How do you integrate lnx / x?

Apr 30, 2016

Use a $u$-substitution to get ${\left(\ln x\right)}^{2} / 2 + C$.

#### Explanation:

At first glance, this integral looks a little confusing because we have a function divided by another function (and those tend to be difficult to work with). But, after rewriting $\int \ln \frac{x}{x} \mathrm{dx}$ as $\int \frac{1}{x} \ln x \mathrm{dx}$, we can see something interesting: we have $\ln x$ and its derivative, $\frac{1}{x}$, in the same integral, making it a textbook case of a $u$-substitution:
Let $\textcolor{b l u e}{u} = \textcolor{b l u e}{\ln} x \to \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \to \textcolor{red}{\mathrm{du}} = \textcolor{red}{\frac{1}{x} \mathrm{dx}}$

Thus, the integral $\int \textcolor{red}{\frac{1}{x}} \textcolor{b l u e}{\ln x} \textcolor{red}{\mathrm{dx}}$ becomes:
$\int \textcolor{b l u e}{u} \textcolor{red}{\mathrm{du}}$

Now, isn't this much easier? Using the reverse power rule, the integral evaluates to ${u}^{2} / 2 + C$. Because $u = \ln x$, we can say:
$\int \ln \frac{x}{x} \mathrm{dx} = {\left(\ln x\right)}^{2} / 2 + C$