How do you evaluate the definite integral \[\int {\log xdx} \] from $\left[ {2,4} \right]$?

Hint: In this question they have given \[\int {\log xdx} \]and asked us to solve it by using parts method. We will have to take any one term as \[u\] and the other as \[dv\] and have to find its derivative and the value of $v$. Then we need to substitute it in the formula \[I = uv - \smallint vdu\] and then simply and solve it to find the correct answer.

Formula used: Formula for integrating in parts method:
\[I = uv - \smallint vdu\]

Complete step by step solution:
In this question they have given \[\int {\log xdx} \] and asked us to solve it by using parts method.
We know that the formula for using part method is \[I = uv - \smallint vdu\]
First we need to decide which one to choose as \[u\] and which one as \[dv\].
Let us take \[u = \log x\] and \[dv = dx.\]
Now, we need to find the derivative of \[u\]and the value of $v$
\[ \Rightarrow u = \log x\]
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\log x$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{x}$
With this we need to find $du$, which is
$ \Rightarrow du = \dfrac{1}{x}dx$
Now we know that \[dv = dx.\]
$ \Rightarrow \dfrac{{dv}}{{dx}} = 1$
$ \Rightarrow v = \int {1dx} $
$ \Rightarrow v = x$
Now, substituting in the formula, \[I = uv - \smallint vdu\]
\[ \Rightarrow \int {\log xdx} = x(\log x) - \smallint \dfrac{1}{x}.xdx\]
Sampling it,
\[ \Rightarrow \int {\log xdx} = x(\log x) - \smallint dx\]
\[ \Rightarrow \int {\log xdx} = x(\log x) - x + C\]
Therefore, the integration of given $I$ is$\int {(\log x)dx = } x\log x - x + C$
Now applying the limit of
\[ \Rightarrow \int\limits_2^4 {\log xdx} = \left( {4\log 4 - 4} \right) - \left( {2(\log 2) - 2} \right)\]
\[ \Rightarrow 4\log {2^2} - 4 - 2\log 2 + 2\]
\[ \Rightarrow 8\log 2 - 2\log 2 - 2\]
Simplifying it, we get
$ \Rightarrow 6\log 2 - 2$

Therefore $6\log 2 - 2$ is the answer.

Note: Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. This method is used to find the integrals by reducing them into standard form. We do not add any constant while finding the integral of the second function.