# How do you integrate ln(3x)?

##### 1 Answer
May 20, 2015

Hello,

Answer : $x \ln \left(3 x\right) - x + c t e$.

Remark that
$\frac{d}{\mathrm{dx}} \left(X \ln \left(X\right) - X\right) = 1. \ln \left(X\right) + \frac{X}{X} - 1 = \ln \left(X\right)$,
therefore
$\int \ln \left(X\right) \mathrm{dX} = X \ln \left(X\right) - X + c t e$

Now, it's easy because, $\ln \left(3 x\right) = \ln \left(3\right) + \ln \left(x\right)$, therefore,
$\int \ln \left(3 x\right) \mathrm{dx} = \ln \left(3\right) x + x \ln \left(x\right) - x + c t e = x \ln \left(3 x\right) - x + c t e$

PS. There is a classical trick to find $\int \ln \left(x\right) \mathrm{dx}$ if you don't have any idea to test $x \ln \left(x\right) - x$ : an integration by parts. The general formula is
$\int u ' v = u v - \int u v '$ (because $\left(u v\right) ' = u ' v + u v '$).
Now, write
$\int 1 \setminus \cdot \ln \left(x\right) \mathrm{dx} = \setminus \int u ' v$, with $u = x$ and $v = \ln \left(x\right)$. You get
$\int \ln \left(x\right) \mathrm{dx} = x \ln \left(x\right) - \setminus \int \frac{x}{x} \mathrm{dx} = x \ln \left(x\right) - x + c t e$