# How do you integrate ln(2x)?

Nov 24, 2016

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

#### Explanation:

$I = \int \ln \left(2 x\right) \mathrm{dx}$

We should use integration by parts in the absence of all other possible integration strategies. Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. For $\int \ln \left(2 x\right) \mathrm{dx}$, let:

$\left\{\begin{matrix}u = \ln \left(2 x\right) & \implies & \mathrm{du} = \frac{2}{2 x} = \frac{1}{x} \\ \mathrm{dv} = \mathrm{dx} & \implies & v = x\end{matrix}\right.$

Thus:

$I = u v - \int v \mathrm{du} = x \ln \left(2 x\right) - \int x \frac{1}{x} \mathrm{dx}$

$I = x \ln \left(2 x\right) - \int \mathrm{dx}$

$I = x \ln \left(2 x\right) - x$

$I = x \left(\ln \left(2 x\right) - 1\right) + C$