Question

# Evaluate the given integral: $\int{{{x}^{x}}\ln \left( ex \right)dx}$ .

Hint: Start by simplification of the integral by using the formula ln(ab)=lna+lnb . After simplification, let the ${{x}^{x}}$ be t and solve the integral.

Before starting the solution, let us discuss the important formulas required for the question.
Some important formulas are:
\begin{align} & \ln ab=\ln a+\ln b \\ & \ln e=1 \\ \end{align}
Now let us start with the integral given in the above question.
$\int{{{x}^{x}}\ln \left( ex \right)dx}$
Now we will use the formula ln(ab)=lna + lnb. On doing so, we get
$\int{{{x}^{x}}\left( \ln e+\ln x \right)dx}$
We know that lne=1.
$\therefore \int{{{x}^{x}}\left( 1+\ln x \right)dx}$
Now to convert the integral to a form from where we can directly integrate it, we let ${{x}^{x}}$ to be t.
${{x}^{x}}=t$
Now if we take log on both the sides of the equation, we get
$x\ln x=\ln t$
$\Rightarrow \ln x+x\times \dfrac{1}{x}=\dfrac{1}{t}\times \dfrac{dt}{dx}$
$\Rightarrow t\left( \ln x+1 \right)=\dfrac{dt}{dx}$
$\Rightarrow {{x}^{x}}\left( 1+\ln x \right)dx=dt$

So, if we substitute the terms in the integral in terms of t, our integral becomes:
$\int{dt}$
Now we know that $\int{dx}$ is equal to x + c. So, our integral comes out to be:
$t+c$
Now we will substitute the value of t as assumed by us to convert our answer in terms of x.
${{x}^{x}}+c$
So, we can say that the value of $\int{{{x}^{x}}\ln \left( ex \right)dx}$ is equal to ${{x}^{x}}+c$ .

Note: Don’t forget to substitute the assumed variable in your integrated expression to reach the final answer. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well.