Question

Solve the following integration:
$\int {{x^x}\log \left( {ex} \right)dx} $
\[
  A.{\text{ }}{x^x} \\
  B.{\text{ }}x\log x \\
  C.{\text{ }} - x{e^{{{\tan }^{ - 1}}x}} \\
  D.{\text{ }}x{e^{{{\tan }^{ - 1}}x}} \\
 \]

Answer 1

Hint: In order to solve the integration first we will simplify the logarithm term. This Integration can be solved after some substitution of variables in place of particular terms which is related to others in some respects. We will substitute the variable in place of the exponential term and would find out the other term respectively to solve the integration.

Complete step by step answer:
We have to solve the integration: $\int {{x^x}\log \left( {ex} \right)dx} $
Let us first simplify the internal term $\log \left( {ex} \right)$
We know that the product of term in logarithm is given as:
$\log \left( {ab} \right) = \log \left( a \right) + \log \left( b \right)$
Using the same formula we have:
$\log \left( {ex} \right) = \log \left( e \right) + \log \left( x \right)$
We know that the value of $\log \left( e \right) = 1$
So, the term becomes:
$ \Rightarrow \log \left( {ex} \right) = 1 + \log \left( x \right)$
For solving the integration $\int {{x^x}\left( {1 + \log \left( x \right)} \right)dx} $
Let us substitute p in the place of ${x^x}$
So, we have:
$p = {x^x}$
Let us take logarithm on both the side
$
   \Rightarrow \log p = \log \left( {{x^x}} \right) \\
   \Rightarrow \log p = x\log \left( x \right) \\
 $
Now let us differentiate the equation with respect to variable p.
\[
   \Rightarrow \dfrac{d}{{dp}}\log p = \dfrac{d}{{dp}}\left[ {x\log \left( x \right)} \right] \\
   \Rightarrow \dfrac{1}{p} = \dfrac{d}{{dx}}\left[ {x\log \left( x \right)} \right] \cdot \dfrac{{dx}}{{dp}} \\
   \Rightarrow \dfrac{1}{p} = \left( {1 + \log x} \right) \cdot \dfrac{{dx}}{{dp}} \\
   \Rightarrow dp = p\left( {1 + \log x} \right) \cdot dx \\
   \Rightarrow dp = {x^x}\left( {1 + \log x} \right) \cdot dx \\
 \]
Let us now substitute the values back to the integration term.
$
   \Rightarrow \int {{x^x}\left( {1 + \log \left( x \right)} \right)dx} = \int {1.dp} \\
   \Rightarrow \int {1.dp} = p \\
 $
So, we have the integration in terms of variable p.
Let us re substitute the variables back in terms of x to get the final answer.
$\int {{x^x}\left( {1 + \log \left( x \right)} \right)dx} = p = {x^x}$
Hence, the integration of the given term is $\int {{x^x}\log \left( {ex} \right)dx} = {x^x}$
So, option A is the correct answer.

Note: In order to solve such types of problems, students should not only try to solve the problem conventionally but rather try to bring out some substitution to reduce the problem as in the case above. But in order to proceed further students must remember different formulas of integration, differentiation and logarithm.