Question

Integrate the following
\[\int{{{x}^{x}}\ln \left( ex \right)dx}\] is equal to: -
(a) \[{{x}^{x}}+C\]
(b) \[x-\ln x+C\]
(c) \[{{\left( \ln x \right)}^{x}}+C\]
(d) \[{{x}^{\ln x}}+C\]

Answer 1

Hint: First of all use the product to sum conversion rule of logarithm given as \[\log \left( ab \right)=\log a+\log b\], to convert \[\ln \left( ex \right)\]. Now, assume \[{{x}^{x}}=k\] and take \[\ln \] (natural log) both sides and use the formula \[\log {{m}^{n}}=n\log m\] to simplify. Differentiate both sides using product rule of differentiation in the L.H.S given as \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]. Use substituting method to replace the variable x with variable k in the integral expression and hence integrate the simplified expression. At last convert the variable k again in variable x after integration.

Complete step by step answer:
Here, we have been provided with the integral \[\int{{{x}^{x}}\ln \left( ex \right)dx}\] and we have to find its value.
Now, let us assume the given integral as I. So, we have,
\[\Rightarrow I=\int{{{x}^{x}}\ln \left( ex \right)dx}\]
Applying the product to sum conversion rule of logarithm given as \[\log \left( ab \right)=\log a+\log b\], we have,
\[\Rightarrow I=\int{{{x}^{x}}\left( \ln e+\ln x \right)dx}\]
Here, \[\ln \] is log to the base ‘e’ and it is called natural logarithm. We know that if the base and argument of a logarithmic expression are equal then its value is 1, i.e. \[{{\log }_{a}}a=1\]. So, we have,
\[\Rightarrow I=\int{{{x}^{x}}\left( 1+\ln x \right)dx}\] - (1)
Now, we are assuming \[{{x}^{x}}=k\], so taking natural \[\log \left( \ln \right)\] both the sides, we have,
\[\Rightarrow \ln {{x}^{x}}=\ln k\]
Using the formula: - \[\log {{m}^{n}}=n\log m\], we have,
\[\Rightarrow x\ln x=\ln k\]
Differentiating both sides with respect to x, we get,
\[\Rightarrow \dfrac{d\left( x\ln x \right)}{dx}=\dfrac{d\left( \ln k \right)}{dx}\]
Using the product rule of differentiation given as, \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] in the L.Hs. and chain rule of differentiation given as, \[\dfrac{df\left( k \right)}{dx}=\left( \dfrac{df\left( k \right)}{dk}\times \dfrac{dk}{dx} \right)\] in the R.H.S, we get,
\[\begin{align}
  & \Rightarrow x\dfrac{d\left( \ln x \right)}{dx}+\ln x\dfrac{dx}{dx}=\dfrac{d\left( \ln k \right)}{dk}\times \dfrac{dk}{dk} \\
 & \Rightarrow x\times \dfrac{1}{x}+\left( \ln x \right)\times 1=\dfrac{1}{k}\times \dfrac{dk}{dx} \\
 & \Rightarrow 1+\ln x=\dfrac{1}{k}\dfrac{dk}{dx} \\
\end{align}\]
Multiplying (dx) on both sides, we get,
\[\Rightarrow \left( 1+\ln x \right)dx=\dfrac{dk}{k}\]
Now, on substituting the value of above obtained expression in equation (i), we get,
\[\begin{align}
  & \Rightarrow I=\int{k\times \dfrac{1}{k}dk} \\
 & \Rightarrow I=\int{dk} \\
 & \Rightarrow I=k+C \\
\end{align}\]
Substituting the assumed value of k again, we get,
\[\Rightarrow I={{x}^{x}}+C\]
Hence, option (a) is the correct answer.

Note:
One may note that here we were provided with the product of two functions \[{{x}^{x}}\] and \[\ln \left( ex \right)\] yet we have not applied the method of integration by parts using the ILATE rule. This is because this method will make our function more difficult to integrate. You must know all the methods of integration by practicing different types of questions so that at the first attempt the method that will be used can be recognized. Remember the product and chain rules of differentiation and properties of logarithm so that the substitution method can be applied.