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Solve this$\int {{x^x}\log \left( {ex} \right)dx} $?
$
  {\text{A}}{\text{. }}{x^2} + c \\
  {\text{B}}{\text{. }}x \cdot \operatorname{logx} + c \\
  {\text{C}}{\text{.}}{\left( {\log x} \right)^x} + c \\
  {\text{D}}{\text{. }}{x^{\log x}} + c \\
 $

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Last updated date: 13th Jul 2024
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Answer
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Hint-To solve this question we need to understand how to solve the integrals by the method of substitution and also the properties of logarithm. Then we need to find the correct answer among given examples.

Complete step-by-step answer:
According to Question take ${\text{I = }}\int {{{\text{x}}^x}{\text{ log}}\left( {{\text{ex}}} \right)dx} $
Now using the product rule of log
${\text{I = }}\int {{{\text{x}}^{\text{x}}}\left( {{\text{log e + log x}}} \right)dx} $
Put the value$\log {\text{e = 1}}$in above equation
${\text{I = }}\int {{{\text{x}}^{\text{x}}}\left( {1 + {\text{log x}}} \right)dx} $ …. (1)
Let ${\text{p = }}{{\text{x}}^x}$ …. (2)
Now, take log on both sides
${\text{log p = x log x}}$
Now differentiate this equation with respect to x
$
  \dfrac{1}{{\text{p}}}\dfrac{{dp}}{{dx}} = {\text{log x + }}\dfrac{{\text{x}}}{{\text{x}}} \\
  \dfrac{{dp}}{{dx}} = {\text{p}}\left( {1 + {\text{ log x}}} \right) \\
  dp = {\text{p}}\left( {1 + {\text{log x}}} \right)dx \\
 $
Now from equation (2), we get
$dp = {{\text{x}}^{\text{x}}}\left( {{\text{1 + log x}}} \right)dx$
Equating this equation with equation (1)
$
  {\text{I = }}\int {dp} \\
  {\text{I = p + c}} \\
 $
From equation (2)
 ${\text{I = }}{{\text{x}}^{{\text{x }}}}{\text{ + c}}$ , where c is a constant.

Note-In these types of questions we need to apply the product rule of log and solve the question by analysing the options and solving Integration by Substitution.
Product rule of log is ${\log _a}\left( {{\text{xy}}} \right) = {\log _a}{\text{x + lo}}{{\text{g}}_a}{\text{y}}$
Integration by Substitution is a method to solve integrals by setting them up in a special specific way. It is also called ‘u-Substitution’ or ‘The Reverse Chain Rule’.