Question

How do you evaluate the integral of $\int {{x^3}\ln \left( x \right)dx} $ ?

Answer 1

Hint: 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. With the help of integration by parts method, we can find the integral of a product function by dividing it out as two separate entities. Then, we differentiate one entity and integrate the other as per the rule so as to obtain the answer.

Complete step by step solution:
We have, $\int {{x^3}\ln \left( x \right)dx} $
So, we consider the given integral as a new variable.
Consider $I = \int {{x^3}\ln \left( x \right)dx} $
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 $\ln \left( x \right)$as first function and ${x^3}$ as second function, we get
\[I = \left[ {\ln \left( x \right)\int {{x^3}dx} } \right] - \int {\left[ {\dfrac{d}{{dx}}\left( {\ln \left( x \right)} \right)\int {{x^3}dx} } \right]} dx\]
Now, we know that the derivative of $\ln \left( x \right)$ with respect to x is $\left( {\dfrac{1}{x}} \right)$. Also, we know that integral of $\left( {{x^3}} \right)$with respect to x is $\left( {\dfrac{{{x^4}}}{4}} \right)$. So, we get,
\[ \Rightarrow I = \left[ {\ln \left( x \right)\left( {\dfrac{{{x^4}}}{4}} \right)} \right] - \int {\left[ {\left( {\dfrac{1}{x}} \right)\left( {\dfrac{{{x^4}}}{4}} \right)} \right]} dx\]
Simplifying the expression further and cancelling the numerator and denominator of the integral expression. We get,
\[ \Rightarrow I = \left[ {\ln \left( x \right)\left( {\dfrac{{{x^4}}}{4}} \right)} \right] - \dfrac{1}{4}\left( {\int {{x^3}} dx} \right)\]
Now, we know that the integral of $\left( {{x^3}} \right)$with respect to x is $\left( {\dfrac{{{x^4}}}{4}} \right)$. We get,
\[ \Rightarrow I = \ln \left( x \right)\left( {\dfrac{{{x^4}}}{4}} \right) - \left( {\dfrac{{{x^4}}}{{16}}} \right) + c\]

So, the value of integral $\int {{x^3}\ln \left( x \right)dx} $ is \[\ln \left( x \right)\left( {\dfrac{{{x^4}}}{4}} \right) - \left( {\dfrac{{{x^4}}}{{16}}} \right) + c\], where c is any arbitrary constant.

Note: Integration by parts method can be used to solve integrals of various complex functions involving 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.