
Evaluate the integral $\int {x\log x dx} $
A. $\dfrac{{{x^2}}}{4}\left( {2\log x - 1} \right) + c$
B. $\dfrac{{{x^2}}}{2}\left( {2\log x - 1} \right) + c$
C. $\dfrac{{{x^2}}}{4}\left( {2\log x + 1} \right) + c$
D. $\dfrac{{{x^2}}}{2}\left( {2\log x + 1} \right) + c$
Answer
204.6k+ views
Hint: The given integrand is a product of two functions $x$ and $\log x$. Use by parts the rule of integration taking $\log x$ as first function and $x$ as second function to evaluate the integral.
Formula Used:
The by parts rule of the integration of product of two functions $u\left( x \right)$ and $v\left( x \right)$ is $\int {u\left( x \right)v\left( x \right)dx = u\left( x \right)\int {v\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}\left\{ {u\left( x \right)} \right\}\int {v\left( x \right)dx} } \right]} dx} } $, where the functions $u\left( x \right)$ and $v\left( x \right)$ are chosen according to ILATE rule.
Complete step by step solution:
Let $u\left( x \right) = \log x$ and $v\left( x \right) = x$
Then the given integral is
$\int {x\log x dx = \int {\log x \cdot xdx} } $
Use the by parts rule
$\log x\int {\left( x \right)dx - \int {\left\{ {\dfrac{d}{{dx}}\left( {\log x} \right)\int {xdx} } \right\}} } dx.....(i)$
Use the formula $\int {\left( {{x^n}} \right)dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$ and $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
Putting $n = 1$, we get
$\int {xdx = \dfrac{{{x^2}}}{2} + c} $
Now, from $(i)$, we get
$\int {x\log xdx} \\ \Rightarrow \log x\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\left( {\dfrac{1}{x} \cdot \dfrac{{{x^2}}}{2}} \right)} dx\\ \Rightarrow \dfrac{{{x^2}}}{2}\log x - \dfrac{1}{2}\int {xdx} \\ \Rightarrow \dfrac{{{x^2}}}{2}\log x - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right) + c\\ \Rightarrow \dfrac{{{x^2}}}{2}\log x - \dfrac{{{x^2}}}{4} + c\\ \Rightarrow \dfrac{{{x^2}}}{4}\left( {2\log x - 1} \right) + c$
Option ‘A’ is correct
Note: Consider using the ILATE property when selecting the first and second functions, and remember that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. For indefinite integrals, the integration constant must be used.
Formula Used:
The by parts rule of the integration of product of two functions $u\left( x \right)$ and $v\left( x \right)$ is $\int {u\left( x \right)v\left( x \right)dx = u\left( x \right)\int {v\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}\left\{ {u\left( x \right)} \right\}\int {v\left( x \right)dx} } \right]} dx} } $, where the functions $u\left( x \right)$ and $v\left( x \right)$ are chosen according to ILATE rule.
Complete step by step solution:
Let $u\left( x \right) = \log x$ and $v\left( x \right) = x$
Then the given integral is
$\int {x\log x dx = \int {\log x \cdot xdx} } $
Use the by parts rule
$\log x\int {\left( x \right)dx - \int {\left\{ {\dfrac{d}{{dx}}\left( {\log x} \right)\int {xdx} } \right\}} } dx.....(i)$
Use the formula $\int {\left( {{x^n}} \right)dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$ and $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
Putting $n = 1$, we get
$\int {xdx = \dfrac{{{x^2}}}{2} + c} $
Now, from $(i)$, we get
$\int {x\log xdx} \\ \Rightarrow \log x\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\left( {\dfrac{1}{x} \cdot \dfrac{{{x^2}}}{2}} \right)} dx\\ \Rightarrow \dfrac{{{x^2}}}{2}\log x - \dfrac{1}{2}\int {xdx} \\ \Rightarrow \dfrac{{{x^2}}}{2}\log x - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right) + c\\ \Rightarrow \dfrac{{{x^2}}}{2}\log x - \dfrac{{{x^2}}}{4} + c\\ \Rightarrow \dfrac{{{x^2}}}{4}\left( {2\log x - 1} \right) + c$
Option ‘A’ is correct
Note: Consider using the ILATE property when selecting the first and second functions, and remember that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. For indefinite integrals, the integration constant must be used.
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