# How do you integrate int lnx/x by integration by parts method?

Oct 19, 2016

$\int \ln \frac{x}{x} \mathrm{dx} = \frac{1}{2} {\left(\ln x\right)}^{2} + c$

#### Explanation:

The formula for IBP is $\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

Se hopefully we can identify one function which simplifies when differentiated, and the other which simplifies when integrated (or is at least easier to integrate than the original function).

so hopefully it is obvious that we choose:
$u = \ln x$ and $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

so $u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$
and $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x} \implies v = \ln x$

$\therefore \int \ln x \frac{1}{x} \mathrm{dx} = \ln x \ln x - \int \ln x \frac{1}{x} \mathrm{dx} + {c}_{1}$
$\therefore \int \ln \frac{x}{x} \mathrm{dx} = {\left(\ln x\right)}^{2} - \int \ln \frac{x}{x} \mathrm{dx} + {c}_{1}$
$\therefore 2 \int \ln \frac{x}{x} \mathrm{dx} = {\left(\ln x\right)}^{2} + {c}_{1}$
$\therefore \int \ln \frac{x}{x} \mathrm{dx} = \frac{1}{2} {\left(\ln x\right)}^{2} + \frac{1}{2} {c}_{1}$
$\therefore \int \ln \frac{x}{x} \mathrm{dx} = \frac{1}{2} {\left(\ln x\right)}^{2} + c$

Oct 19, 2016

$\implies I = \left(\frac{1}{2}\right) {\left(\ln x\right)}^{2} + K$

#### Explanation:

integration by parts formula:

$\int u \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \mathrm{dx} = u v - \int v \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \mathrm{dx}$

for $\textcolor{red}{I = \left(\int \left(\ln \frac{x}{x}\right) \mathrm{dx}\right)}$

Let $u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

Let $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x} \implies v = \ln x$

$\therefore I = \left(\ln x\right) \left(\ln x\right) - \textcolor{red}{\int \left(\ln \frac{x}{x}\right) \mathrm{dx}}$

$\implies I = {\left(\ln x\right)}^{2} - \textcolor{red}{I} + C$

$\implies 2 I = {\left(\ln x\right)}^{2} + C$

$\implies I = \left(\frac{1}{2}\right) {\left(\ln x\right)}^{2} + K$