Question

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

Answer 1

In this problem, we have to integrate the given integral. We know that, by analysing the above integral, we can use the integration by parts formula to integrate it. We have to use the ILATE method where we can take the first term as the first function in such a way that the first function could be easily integrated. We can then integrate and get the answer.

Complete step by step solution:
We can now write the given integral,
$\Rightarrow \int{\left( {{x}^{3}} \right)\left( \ln x \right)dx}$……… (1)
We know that, by analysing the above integral, we can use the integration by parts formula to integrate it.
We know that the integration by parts formula is,
$\Rightarrow \int{udv=uv-\int{vdu}}$ ……… (2)
We can now use the ILATE rule, where we have a product of two functions.
We can now take the above indefinite integral (1).
We can now use the ILATE method to find which functions to be used in integration by parts.
We know that ILATE stands for Inverse Logarithmic Algebraic Trigonometric Exponential.
We can choose the functions according to the order of letters in ILATE.
We have to take integration for ${{x}^{3}}\ln \left( x \right)$, where $\ln \left( x \right)$ is the logarithmic function and ${{x}^{3}}$ is algebraic function.
Then taking by parts we choose $\ln \left( x \right)$ for L (of ILATE) as first function, u and ${{x}^{3}}$ for A (of ILATE) as second function, $dv$
Such that $u=\ln x$ and $dv={{x}^{3}}dx$
Then, $du=\dfrac{1}{x}dx$ and $v=\dfrac{{{x}^{4}}}{4}$ .
We can substitute the above values in the formula (2), we get
$\Rightarrow \int{\left( {{x}^{3}} \right)\left( \ln x \right)dx=\ln x\left( \dfrac{{{x}^{4}}}{4} \right)-\int{\left( \dfrac{{{x}^{4}}}{4} \right)\left( \dfrac{1}{x} \right)}dx}$
We can now simplify the above term and integrate the terms, we get
$\Rightarrow \int{\left( {{x}^{3}} \right)\left( \ln x \right)dx=\left( \dfrac{{{x}^{4}}\ln x}{4} \right)-\dfrac{1}{4}\int{{{x}^{3}}}dx}$
We can now integrate the above step, we get
$\Rightarrow \dfrac{{{x}^{4}}\ln x}{4}-\dfrac{{{x}^{4}}}{16}+C$
Therefore, the answer is $\dfrac{{{x}^{4}}\ln x}{4}-\dfrac{{{x}^{4}}}{16}+C$.

Note: We should also remember that $du$ is obtained by differentiating u and v is obtained by integrating $dv$, which we use in the integration by parts formula. We know that ILATE stands for Inverse Logarithmic Algebraic Trigonometric Exponential. We can choose the functions according to the order of letters in ILATE.