Question

The value of the integral \[\int{{{x}^{x}}\ln \left( ex \right)dx}\] is equal to:
\[\left( \text{a} \right)\text{ }{{x}^{x}}+C\]
\[\left( \text{b} \right)\text{ }x.\ln x+C\]
\[\left( \text{c} \right)\text{ }{{\left( \ln x \right)}^{x}}+C\]
\[\left( \text{d} \right)\text{ }{{x}^{\ln x}}+C\]

Answer 1

Hint: To solve the given question, we will first consider that the value of \[{{x}^{x}}=t.\] Then we will find the value of dx by differentiating \[{{x}^{x}}=t\] on both sides with respect to x. For differentiating it, we will take a natural logarithm on both sides, i.e. x log x = log t, and then we will differentiate it and find the value of x. Now, we will put the values of \[{{x}^{x}}\] and dx in the integral and solve it.

Complete step by step answer:
We can see that, in the integration, the terms given are very different, i.e. \[{{x}^{x}}\] and ln(ex) are not simple functions. So, we cannot apply by parts to solve the integral. Thus, the only option left for us is to do substitution in the integral. So, we will consider that, \[{{x}^{x}}=t.\] Thus, we have,
\[{{x}^{x}}=t......\left( i \right)\]
Now, we will differentiate both sides of the equation (i) with respect to x. Now there is no direct method of differentiation of the function \[{{x}^{x}}.\] So, we will take a natural logarithm on both sides. Thus we will get,
\[{{\log }_{e}}\left( {{x}^{x}} \right)={{\log }_{e}}t....\left( ii \right)\]
In the above equation, we are going to use a logarithmic identity shown below,
\[\log {{a}^{b}}=b\log a\]
Thus, we will get,
\[x{{\log }_{e}}x={{\log }_{e}}t\]
Now, we will differentiate both sides.
\[\dfrac{d}{dx}\left[ x{{\log }_{e}}x \right]=\dfrac{d}{dx}\left[ {{\log }_{e}}t \right]....\left( iii \right)\]
Now, to find the derivative of \[x{{\log }_{e}}x\] we will use the product rule which says that,
\[\dfrac{d}{dx}\left[ f\left( x \right).g\left( x \right) \right]=g\left( x \right)\left[ \dfrac{d}{dx}f\left( x \right) \right]+f\left( x \right)\left[ \dfrac{d}{dx}g\left( x \right) \right]\]
Thus, we have,
\[\Rightarrow \dfrac{d}{dx}\left[ x{{\log }_{e}}x \right]={{\log }_{e}}x\left[ \dfrac{d}{dx}\left( x \right) \right]+x\left[ \dfrac{d}{dx}\left( {{\log }_{e}}x \right) \right]\]
\[\Rightarrow \dfrac{d}{dx}\left[ x{{\log }_{e}}x \right]={{\log }_{e}}x\left( 1 \right)+x\dfrac{d}{dx}\left( {{\log }_{e}}x \right)\]
The differentiation of log x is \[\dfrac{1}{x}.\] So, we have,
\[\Rightarrow \dfrac{d}{dx}\left[ x{{\log }_{e}}x \right]={{\log }_{e}}x+x\left( \dfrac{1}{x} \right)\]
\[\Rightarrow \dfrac{d}{dx}\left[ x{{\log }_{e}}x \right]={{\log }_{e}}x+1.....\left( iv \right)\]
From (iii) and (iv), we have,
\[\Rightarrow {{\log }_{e}}x+1=\dfrac{d}{dx}\left[ {{\log }_{e}}t \right]\]
\[\Rightarrow {{\log }_{e}}x+1=\dfrac{d}{dt}\times \dfrac{dt}{dx}\left[ {{\log }_{e}}t \right]\]
\[\Rightarrow {{\log }_{e}}x+1=\dfrac{dt}{dx}\left[ \dfrac{d}{dt}{{\log }_{e}}t \right]\]
\[\Rightarrow {{\log }_{e}}x+1=\dfrac{dt}{dx}.\dfrac{1}{t}\]
\[\Rightarrow \left( {{\log }_{e}}x+1 \right)dx=\dfrac{dt}{t}\]
\[\Rightarrow dx=\dfrac{dt}{t\left( {{\log }_{e}}x+1 \right)}.....\left( v \right)\]
Now, we will put the values of \[{{x}^{x}}\] and dx in the integral given in the question.
\[\int{{{x}^{x}}\ln \left( ex \right)dx}=\int{t.\ln \left( ex \right).\dfrac{dt}{t\left( {{\log }_{e}}x+1 \right)}}\]
\[\Rightarrow \int{{{x}^{x}}\ln \left( ex \right)dx}=\int{\dfrac{\ln \left( ex \right)dt}{{{\log }_{e}}x+1}}\]
Now, we will use an exponential identity shown below,
\[{{\log }_{a}}b+1={{\log }_{a}}ab\]
Thus, we will get,
\[\Rightarrow \int{{{x}^{x}}\ln \left( ex \right)dx=\int{\dfrac{\log \left( ex \right)dt}{\log \left( ex \right)}}}\]
\[\Rightarrow \int{{{x}^{x}}\log \left( ex \right)dx}=\int{dt}\]
\[\Rightarrow \int{{{x}^{x}}\log \left( ex \right)dx}=t+C\]
\[\Rightarrow \int{{{x}^{x}}\log \left( ex \right)dx}={{x}^{x}}+C\left[ \text{From }\left( i \right) \right]\]
Hence, option (a) is the right answer.

Note: In the solution, we have used the identity, \[\log {{a}^{b}}=b\log a.\] We cannot use this identity everywhere. To use this identity, a should be greater than 0. Also, for the product rule, the functions f(x) and g(x) should be differentiable in the given domain.