# How do you integrate 2 ln (x-5)?

$\textcolor{b l u e}{\int 2 \ln \left(x - 5\right) \mathrm{dx} = \left(2 x - 10\right) \cdot \ln \left(x - 5\right) - 2 x + C}$

#### Explanation:

The given

$\int 2 \ln \left(x - 5\right) \mathrm{dx}$

Let $u = \ln \left(x - 5\right)$
Let $\mathrm{dv} = \mathrm{dx}$
Let $v = x$
Let $\mathrm{du} = \left(\frac{1}{x - 5}\right) \cdot \mathrm{dx}$
Using integration by parts

$\int u \cdot \mathrm{dv} = u v - \int v \cdot \mathrm{du}$

$\int \ln \left(x - 5\right) \mathrm{dx} = x \cdot \ln \left(x - 5\right) - \int \frac{x}{x - 5} \mathrm{dx}$

$\int \ln \left(x - 5\right) \mathrm{dx} = x \cdot \ln \left(x - 5\right) - \int \frac{x - 5 + 5}{x - 5} \mathrm{dx}$

$\int \ln \left(x - 5\right) \mathrm{dx} = x \cdot \ln \left(x - 5\right) - \int \left(1 + \frac{5}{x - 5}\right) \mathrm{dx}$

$\int \ln \left(x - 5\right) \mathrm{dx} = x \cdot \ln \left(x - 5\right) - x - 5 \cdot \ln \left(x - 5\right)$

So that

$\int 2 \ln \left(x - 5\right) \mathrm{dx} = 2 \cdot \left[x \cdot \ln \left(x - 5\right) - x - 5 \cdot \ln \left(x - 5\right)\right]$

$\int 2 \ln \left(x - 5\right) \mathrm{dx} = 2 x \cdot \ln \left(x - 5\right) - 2 x - 10 \cdot \ln \left(x - 5\right) + C$

$\textcolor{b l u e}{\int 2 \ln \left(x - 5\right) \mathrm{dx} = \left(2 x - 10\right) \cdot \ln \left(x - 5\right) - 2 x + C}$

God bless....I hope the explanation is useful.