Question

Find the integral of $\int {\left( {x + 1} \right)} \log xdx$

Answer 1

Hint: To solve this problem, we will first separate the main term into two terms by opening the bracket. Then, we need to use the integration by parts method for the integration of both the terms. For that, we will first decide which term can be taken as u and which can be taken as v. Here, we will take logarithmic function as u in while doing integration by parts.
Formula used:
Integration by parts of two terms u and v is given by:
 \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx\]
Where, \[u\] is the function \[u\left( x \right)\] , \[v\] is the function \[v\left( x \right)\] and \[u'\] is the derivative of function \[u\left( x \right)\] .

Complete step-by-step answer:
We are asked to find the integration of $\int {\left( {x + 1} \right)} \log xdx$
$I = \int {\left( {x + 1} \right)} \log xdx$
We will separate the main term into two terms by opening the bracket.
$I = \int x \log xdx + \int {1 \cdot \log xdx} $
Thus, we now have two integrations, \[I = {I_1} + {I_2}\]
${I_1} = \int {x\log xdx} $
Here, we will take $u = \log x$and $v = x$.
Now, we will apply the integration by parts rule \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx\]
$
   \Rightarrow {I_1} = \log x\int {xdx} - \int {\left( {\dfrac{{d\left( {\log x} \right)}}{{dx}}\int {xdx} } \right)} dx \\
   \Rightarrow {I_1} = \log x\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\left( {\dfrac{1}{x} \cdot \dfrac{{{x^2}}}{2}} \right)} dx \\
   \Rightarrow {I_1} = \dfrac{{{x^2}\log x}}{2} - \int {\dfrac{x}{2}} dx \\
   \Rightarrow {I_1} = \dfrac{{{x^2}\log x}}{2} - \dfrac{{{x^2}}}{4} + {c_1} \;
 $
Here, ${c_1}$is the constant of integration.
${I_2} = \int {1 \cdot \log xdx} $
Here, we will take $u = \log x$and $v = 1$.
Now, we will apply the integration by parts rule \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx\]
 \[
   \Rightarrow {I_2} = \log x\int {1dx} - \int {\left( {\dfrac{{d\left( {\log x} \right)}}{{dx}}\int {1dx} } \right)} dx \\
   \Rightarrow {I_2} = \log x \cdot x - \int {\left( {\dfrac{1}{x} \cdot x} \right)} dx \\
   \Rightarrow {I_2} = x\log x - \int 1 dx \\
   \Rightarrow {I_2} = x\log x - x + {c_2} \;
 \]
Here, ${c_2}$is the constant of integration.
We know that
 \[
  I = {I_1} + {I_2} \\
   \Rightarrow I = \dfrac{{{x^2}\log x}}{2} - \dfrac{{{x^2}}}{4} + {c_1} + x\log x - x + {c_2} \\
   \Rightarrow I = \dfrac{{{x^2}\log x}}{2} - \dfrac{{{x^2}}}{4} + x\log x - x + {c_1} + {c_2} \\
   \Rightarrow I = \dfrac{{{x^2}\log x}}{2} - \dfrac{{{x^2}}}{4} + x\log x - x + c \;
 \]
Where, $c = {c_1} + {c_2}$ is the constant of integration.
Thus, our final answer is: $\int {\left( {x + 1} \right)} \log xdx = \dfrac{{{x^2}\log x}}{2} - \dfrac{{{x^2}}}{4} + x\log x - x + c$
So, the correct answer is “$\dfrac{{{x^2}\log x}}{2} - \dfrac{{{x^2}}}{4} + x\log x - x + c$
”.

Note: Here, we have used integration by parts method which is a special method of integration that is often useful when two functions are multiplied together. In this rule it is very important to choose write functions as u and v. There is a helpful rule to remember for choosing which function to take as u. This rule is called ILATE rule, where I stands for inverse trigonometric functions, L stands for logarithmic functions, A stands for algebraic functions, T stands for trigonometric functions and E stands for exponential functions.