Integrate the function $ x\log 2x $ .

Hint: In this problem we will apply the method of integration by parts. The method of integration by parts is generally used when we have to integrate the product of two functions. As per this method we take the functions as u and v and then proceed. We use the ILATE rule to choose u and v.

Complete step-by-step answer:
Let us represent the given function by f(x). So, we have:
 $ f\left( x \right)=x\log 2x $
For integrating this function the best method to be used is integration by parts.
In calculus, integration by parts or partial integration is the method which is mostly used to find integration of the product of the two functions.
In this one of the function is taken to be u and other as v then the integration of the function $ u.v $ with respect to x is given as:
 $ \int_{{}}^{{}}{u.vdx=u.\int_{{}}^{{}}{vdx-\int_{{}}^{{}}{\dfrac{du}{dx}\int_{{}}^{{}}{v.dx...........(1)}}}} $
So, for the given function according to the ILATE rule, we will take u=log2x and v = x.
On substituting the values of u and v in equation 1, we get:
\[\int_{{}}^{{}}{x\log 2xdx=\log 2x\int_{{}}^{{}}{xdx-\int_{{}}^{{}}{\dfrac{d\left( \log 2x \right)}{dx}\int_{{}}^{{}}{xdx}}}}........\left( 2 \right)\]
We also have:
 $ \int_{{}}^{{}}{xdx=\dfrac{{{x}^{2}}}{2}} $ and $ \dfrac{d\left( \log 2x \right)}{dx}=\dfrac{1}{2x}\times \dfrac{d\left( 2x \right)}{dx}=\dfrac{1}{2x}\times 2=\dfrac{1}{x} $
So, on putting these values in equation (2), we get:
\[\begin{align}
  & \int_{{}}^{{}}{x\log 2xdx=\log 2x\times \dfrac{{{x}^{2}}}{2}-\int_{{}}^{{}}{\dfrac{1}{x}\times \dfrac{{{x}^{2}}}{2}}} \\
 & \Rightarrow \int_{{}}^{{}}{x\log 2xdx=}\log 2x\times \dfrac{{{x}^{2}}}{2}-\int_{{}}^{{}}{\dfrac{x}{2}dx} \\
 & \Rightarrow \int_{{}}^{{}}{x\log 2xdx=}\dfrac{1}{2}{{x}^{2}}\log 2x-\dfrac{1}{2}\times \dfrac{{{x}^{2}}}{2}+c \\
 & \Rightarrow \int_{{}}^{{}}{x\log 2xdx=}\dfrac{{{x}^{2}}\log 2x}{2}-\dfrac{{{x}^{2}}}{4}+c \\
\end{align}\]
Here, c is the constant of integration.
Hence, the integration of the function $ x\log 2x $ is \[\int_{{}}^{{}}{x\log 2xdx=\dfrac{{{x}^{2}}\log 2x}{2}}-\dfrac{{{x}^{2}}}{4}+c\].

Note: While choosing u and v functions, students should keep the ILATE rule in mind. In ILATE, as we know that I stands for inverse logarithmic function, L stands for logarithmic function, A stands for algebraic function, T stands for trigonometric function and A stands for algebraic function. Here I come before A, hence we give first preference to L.