# How do you evaluate the integral of ln(2x)/x^2 dx?

Mar 16, 2015

You can evaluate this integral by using integration by parts.

Let $u = \ln \left(2 x\right)$

Let $\mathrm{dv} = \frac{1}{x} ^ 2 \mathrm{dx}$

Differentiating $u = \ln \left(2 x\right)$ we have

$\mathrm{du} = \frac{1}{x} \mathrm{dx}$

Integrating $\mathrm{dv} = \frac{1}{x} ^ 2 \mathrm{dx}$ we have

$\int \mathrm{dv} = \int {x}^{-} 2 \mathrm{dx}$

$v = - \frac{1}{x}$

Recall the integration by parts formula

$u v - \int v \mathrm{du}$

Now proceed as follows

$- \ln \frac{2 x}{x} - \int - \frac{1}{x} \left(\frac{1}{x}\right) \mathrm{dx}$

$- \ln \frac{2 x}{x} - \int - \frac{1}{x} ^ 2 \mathrm{dx}$

$- \ln \frac{2 x}{x} + \int {x}^{-} 2 \mathrm{dx}$

$- \ln \frac{2 x}{x} - \frac{1}{x} + C$

We can rewrite if you like

$- \frac{\left(\ln \left(2 x\right) + 1\right)}{x} + C$

General note:
For $n \ne - 1$, any integral of the form $\int {x}^{n} \ln x \mathrm{dx}$ can be found by the process described above.

(For $n = - 1$ use $u$-substitution, with $u = \ln x$ so $\mathrm{du} = \frac{1}{x} \mathrm{dx}$.)