Question

The value of $\int {\log \left( x \right)dx}$ is
A. $x\log \left( {\dfrac{x}{e}} \right) + c$
B. ${x^2}\log (x) + c$
C. ${e^x}\log (x) + c$
D. $(x + 1)\log (x) + c$

Answer 1

this is a question of indefinite integration. We can use by-parts integration. We can use part integration to solve this integral. We can use log(x) as the first function and $1$ as the second function. Also, we can assume this integral to be equal to $I$, and then we can proceed with our calculation.

Formula used:
$\int {uvdx = \left[ {u\int {vdx - \int {u'\left( {\int {vdx} } \right)dx} } } \right]}$

Complete step by step answer:
let the given integral be equal to $I$.
$I = \int {\log xdx}$
Using by-parts integration,
Take the first function as $\log x$ and the second function as $1$.
$I = \left[ {\log (x)\int {1dx} - \int {\dfrac{{d\left( {\log x} \right)}}{{dx}} \times \left( {\int {1dx} } \right)dx} } \right]$
Integration of $1$ is x.
$I = \left[ {\log x\left( x \right) - \int {\dfrac{{d\left( {\log x} \right)}}{{dx}} \times xdx} } \right]$
Differentiation of log(x) is $\dfrac{1}{x}$.
$I = \,\left[ {\log \left( x \right) - \int {\dfrac{x}{x}dx} } \right]$
On dividing, we get
I = \left[ {\log \left( x \right) - \int {1dx} } \right] \\\
\Rightarrow I = x\log \left( x \right) - x + c \\\
Here, c is the constant of integration. We can also write $1$ as log(e).Therefore,
I = \left[ {x\log (x) - x\log (e) + c} \right] \\\
Taking x common,
$I = x\left[ {\log \left( x \right) - \log \left( e \right)} \right] + c \\\$
Now, using the property of logarithm,
\log \left( {\dfrac{a}{b}} \right) = \log \left( a \right) - \log \left( b \right) \\\
$\therefore I = x\log \left( {\dfrac{x}{e}} \right) + c$

Hence, the correct answer is option A.

Note: Using $1$ as a function in by-parts integration is a standard format of integration. There is one alternative to this question. You can also differentiate the options and after differentiation, the option which matches with the integral will be the correct answer. On differentiation, the constant of integration becomes zero. We can use this method in simple questions but not in complex ones.