Question

The \[\int {x\log xdx} \] equals:
A. $\left( {\dfrac{{{x^2}}}{4}} \right)\left[ {2\log x - 1} \right] + c$
B. $\left( {\dfrac{{{x^2}}}{2}} \right)\left[ {2\log x - 1} \right] + c$
C. $\left( {\dfrac{{{x^2}}}{4}} \right)\left[ {2\log x + 1} \right] + c$
D. $\left( {\dfrac{{{x^2}}}{2}} \right)\left[ {2\log x + 1} \right] + c$

Answer 1

Hint: In the given question, we have to find the indefinite integral of the function \[x\log x\] with respect to x. We will use the integration by parts method to evaluate the given integral. We will set the order of functions according to the acronym ILATE. We must remember the integral of some basic functions in order to answer the problem.

Complete step by step answer:
In the given question, we are required to find the value of the integral given to us in the question by using the integration by parts method. So, we consider the given integral as a new variable. Consider $I = \int {x\log xdx} $. In integration by parts method, we integrate a function which is a product of two functions using a formula:
$\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } $

Hence, using integration by parts method and considering $\log x$ as first function and $x$ as second function, we get
\[I = \left[ {\log x\int {xdx} } \right] - \int {\left[ {\dfrac{d}{{dx}}(\log x).\int {xdx} } \right]} dx\]
Now, we know that the derivative of logarithmic function with respect to x is $\dfrac{1}{x}$. Also, we know that the integral of $x$with respect to x is $\left( {\dfrac{{{x^2}}}{2}} \right)$.
\[ \Rightarrow I = \left[ {\log x \times \left( {\dfrac{{{x^2}}}{2}} \right)} \right] - \int {\left[ {\dfrac{1}{x} \times \left( {\dfrac{{{x^2}}}{2}} \right)} \right]} dx\]
\[ \Rightarrow I = \left[ {\log x \times \left( {\dfrac{{{x^2}}}{2}} \right)} \right] - \int {\dfrac{x}{2}\,} dx\]

Taking constant out of the integral, we get,
\[ \Rightarrow I = \left( {\dfrac{{{x^2}}}{2}} \right)\log x - \dfrac{1}{2}\int {x\,} dx\]
Again using the power rule of integration $\int {{x^n}dx} = \left( {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right) + c$. So, we get,
\[ \Rightarrow I = \left( {\dfrac{{{x^2}}}{2}} \right)\log x - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right) + c\], where c is any arbitrary constant.
\[ \Rightarrow I = \dfrac{{{x^2}}}{2}\log x - \dfrac{{{x^2}}}{4} + c\]
Now, factoring out the common factors, we get,
\[ \therefore I = \dfrac{{{x^2}}}{4}\left( {2\log x - 1} \right) + c\]
So, the integral \[\int {x\log xdx} \] equals \[\dfrac{{{x^2}}}{4}\left( {2\log x - 1} \right) + c\].

Hence, the correct answer is option B.

Note: Integration by parts method can be used to solve integrals of various complex functions involving the product of functions within itself easily. In the given question, the use of integration by parts method once has made the process much easier and organized. However, this might not be the case every time. We may have to apply the integration by parts method multiple times in order to reach the final answer. We must not forget to add c to the answer of indefinite integral as it represents a family of curves.