Question

The value of $\int {\log }_{}\left(\text{x}\right)\text{dx}$ is

Answer 1

We can write $\text{log(x)=1.log(x)}$ so

$\int {\log }_{}\left(\text{x}\right)\text{dx=}$$\int 1.{\log }_{}\left(x\right)\text{dx}$

We can solve it by using integration by parts ,

For this we take $\text{log(x)}$as first function and $1$ as second function .

$$\begin{gathered} \int 1.\log \left(x\right)dx=\log \left(x\right)\int 1.dx-\int \left(\frac{d}{dx}\right(\log x)\int 1.dx) \\ =\log \left(x\right).x-\int \frac{1}{x}\times xdx \\ =\log \left(x\right).x-\int 1.dx \\ =\log \left(x\right)x-x+c \\ =x\left(\log \right(x)-1)+c \end{gathered}$$

Hence, value of $\int {\log }_{}\left(\text{x}\right)\text{dx}$ is $x\left(\log \right(x)-1)+c$.