Question

How to integrate $\ln \left(x\right)$?

Answer 1

Integrate the given expression :

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

We will use the method of by parts to solve the integration.

We now that

$\int udv=uv-\int vdu$

Let ,

1. $u=\ln \left(x\right)$
2. $dv=dx$

$\Rightarrow v=x$

Therefore,

$$\begin{gathered} \mathrm{I}=\int \ln \left(x\right)dx \\ =x\ln \left(x\right)-\int \frac{d\ln \left(x\right)}{dx}·xdx \\ =x\ln \left(x\right)-\int \frac{1}{x}·xdx\left[\because \frac{d\ln \left(x\right)}{dx}=\frac{1}{x}\right] \\ =x\ln \left(x\right)-\int dx \\ =x\ln \left(x\right)-x+c \end{gathered}$$

$\therefore \int \ln \left(x\right)dx=x\ln \left(x\right)-x+c$

Hence, the value of the integrating $\int \ln \left(x\right)dx$ is $x\ln \left(x\right)-x+c$