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Question: The value of integral \(\int{{{x}^{3}}\log x dx}\) is \[\] A.\(\dfrac{1}{16}\left( 4{{x}^{4}}\log ...

The value of integral ∫x3log⁡xdx\int{{{x}^{3}}\log x dx} is A.$\dfrac{1}{16}\left( 4{{x}^{4}}\log x-{{x}^{4}}+c \right)$
B. 18(x4log⁡x−4x4+c)\dfrac{1}{8}\left( {{x}^{4}}\log x-4{{x}^{4}}+c \right) C. $\dfrac{1}{16}\left( 4{{x}^{4}}\log x+{{x}^{4}}+c \right)$
D. x4log⁡x4+c\dfrac{{{x}^{4}}\log x}{4}+c $$$$

Explanation

Solution

We see that in the integrand of the given integral two functions one algebraic x3{{x}^{3}} and one logarithmic log⁡x\log x are multiplied to each other. We integrate by parts first by taking the logarithmic function as fist function f(x)=log⁡xf\left( x \right)=\log x and algebraic function as second function g(x)=x3g\left( x \right)={{x}^{3}} following the ILATE rule and then using the standard form of integration by parts ∫(f(x)g(x))dx=f(x)∫g(x)dx−∫(f′(x)∫g(x)dx)dx\int{\left( f\left( x \right)g\left( x \right) \right)}dx=f\left( x \right)\int{g\left( x \right)dx-\int{\left( {{f}^{'}}\left( x \right)\int{g\left( x \right)dx} \right)}}dx. $$$$

Complete step-by-step solution:
We know that if there two single variable real valued integrable functions in product form say ff and gg then we integrate them by parts taking ff as first function and gg as second function using the formula
∫(f(x)g(x))dx=f(x)∫g(x)dx−∫(f′(x)∫g(x)dx)dx\int{\left( f\left( x \right)g\left( x \right) \right)}dx=f\left( x \right)\int{g\left( x \right)dx-\int{\left( {{f}^{'}}\left( x \right)\int{g\left( x \right)dx} \right)}}dx
The choice of first function depends upon how many times differentiating the function will make the function zero. So the rule that is used when we are integrating by parts is called ILATE, an acronym for inverse, logarithm, algebraic, trigonometric and finally exponent. It means we have to choose the first function in the order of ILATE. $$$$
We know the standard integral of polynomial xn{{x}^{n}} for some n≠−1,n∈Rn\ne -1,n\in R as
∫xndx=xn+1n+1+c\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c
Here cc is a real constant of integration. We know the standard derivative of logarithmic function
ddxlog⁡x=1x\dfrac{d}{dx}\log x=\dfrac{1}{x}
We are asked in the question to evaluate the integral
I=∫x3log⁡xdxI=\int{{{x}^{3}}\log x dx}
We observe the integrand of the integral x3log⁡x{{x}^{3}}\log x has algebraic function x3{{x}^{3}} and the logarithmic function log⁡x\log x. So we follow the rule of ILATE and choose the logarithmic function as first function f(x)=log⁡xf\left( x \right)=\log x and chose the algebraic function as the second function g(x)=x3g\left( x \right)={{x}^{3}}. We use integration parts to have,

& I=\int{{{x}^{3}}\log x dx} \\\ & \Rightarrow I=\log x\int{{{x}^{3}}dx}-\int{\dfrac{d}{dx}\left( \log x \right)\int{{{x}^{3}}dx}} \\\ \end{aligned}$$ We use the standard integration $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for $n=3$ and proceed to have, $$\begin{aligned} & I=\int{{{x}^{3}}\log x dx} \\\ & \Rightarrow I=\log x\left( \dfrac{{{x}^{3+1}}}{3+1} \right)-\int{\dfrac{1}{x}}\left( \dfrac{{{x}^{3+1}}}{3+1} \right)dx \\\ & \Rightarrow I=\log x\left( \dfrac{{{x}^{4}}}{4} \right)-\int{\dfrac{1}{x}}\times \dfrac{{{x}^{4}}}{4}dx \\\ & \Rightarrow I=\log x\left( \dfrac{{{x}^{4}}}{4} \right)-\dfrac{1}{4}\int{{{x}^{3}}dx} \\\ \end{aligned}$$ We use the standard integration $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for $n=3$ and proceed to have, $$\begin{aligned} &\Rightarrow I=\log x\left( \dfrac{{{x}^{4}}}{4} \right)-\dfrac{1}{4}\left( \dfrac{{{x}^{3+1}}}{3+1} \right)+{{c}_{1}} \\\ & \Rightarrow I=\dfrac{{{x}^{4}}\log x}{4}-\dfrac{{{x}^{4}}}{16}+{{c}_{1}} \\\ & \Rightarrow I=\dfrac{1}{16}\left( 4{{x}^{4}}\log x-{{x}^{4}}+c \right) \\\ \end{aligned}$$ **Here $c=16{{c}_{1}}$ is a real constant of integration. So the value of integral is $\dfrac{1}{16}\left( 4{{x}^{4}}\log x-{{x}^{4}}+c \right)$ and the correct option is C.** **Note:** We note that we can also choose the algebraic function as the first function but the integration will require successive use of by parts integration. We can alternatively evaluate the integral using repeated integration $\int{fg}={{f}_{1}}{{g}^{1}}-{{f}_{2}}{{g}^{2}}+{{f}_{3}}{{g}^{3}}...$ where ${{f}_{i}}$ is the ${{i}^{\text{th}}}$ successive integration of $f$ and ${{g}^{i}}$ is the ${{i}^{\text{th}}}$ successive differentiation of $g$. The ILATE rule is also called the LIATE rule in some text books and both rules are valid.