Question

Evaluate the following integral \[\int{{{\left( \log x \right)}^{2}}dx}\] .
(A) \[x\left[ {{\left( \log x \right)}^{2}}-2\log x+2 \right]\]
(B) \[x\left[ {{\left( \log x \right)}^{2}}-2\log x+1 \right]\]
(C) \[x\left[ {{\left( \log x \right)}^{2}}-2\log x-2 \right]\]
(D) \[x\left[ {{\left( \log x \right)}^{2}}+2\log x-2 \right]\]

Answer 1

Hint: Let us assume that \[u=v=\left( \log x \right)\] . We know the integration by parts formula, \[\int{uvdx=u\int{vdx-\left[ \int{\left\{ \dfrac{du}{dx}\int{vdx} \right\}dx} \right]}}\] . Now, use integration by parts formula and simplify the given expression \[\int{{{\left( \log x \right)}^{2}}dx}\] . Use the formula, \[\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}\] and simplify it further. Now, the value of \[\int{\log xdx}\] is required. Assume that \[u=1\] and \[v=\log x\] . Now, apply integration by parts formula and calculate the value of \[\int{\log xdx}\] . Using the value of \[\int{\log xdx}\] , solve the expression \[\int{{{\left( \log x \right)}^{2}}dx}\] further and get its value.

Complete step by step answer:
According to the question, we are given an expression and we have to integrate that expression with respect to \[dx\].
The given expression = \[\int{{{\left( \log x \right)}^{2}}dx}=\int{\left( \log x \right)\left( \log x \right)dx}\] …………………………………….(1)
We know the formula for integration by parts \[\int{uvdx=u\int{vdx-\left[ \int{\left\{ \dfrac{du}{dx}\int{vdx} \right\}dx} \right]}}\] …………………………………………..(2)
For equation (1), let us assume that \[u=v=\left( \log x \right)\] .
Now, using equation (2) and on simplifying equation (1), we get
\[\int{\left( \log x \right)\left( \log x \right)dx=\log x\int{\left( \log x \right)dx-\left[ \int{\left\{ \dfrac{d\left( \log x \right)}{dx}\int{\left( \log x \right)dx} \right\}dx} \right]}}\] ………………………………….(3)
To simplify the above equation, we know the formula, \[\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}\] …………………………………….(4)
Now, from equation (3) and equation (4), we get
\[=\log x\int{\left( \log x \right)dx-\left[ \int{\left\{ \dfrac{1}{x}\int{\left( \log x \right)dx} \right\}dx} \right]}\] ………………………………….(5)
We can observe that our solution is struck and it can be solved easily. That is, for solving equation (5), we need the value of \[\int{\log xdx}\] .
\[=\int{\left( \log x \right).1dx}\] ……………………………………………(6)
Here, in the above equation, let us assume that \[u=1\] and \[v=\log x\] .
Now, from equation (2) and equation (6), we get
 \[=\int{\left( \log x \right).1dx}\]
\[=\log x\int{1dx-\left[ \int{\left\{ \dfrac{d\left( \log x \right)}{dx}\int{1dx} \right\}dx} \right]}\] …………………………………………..(7)
Now, from equation (4) and equation (7), we get
\[\begin{align}
  & =x\log x-\left[ \int{\left\{ \dfrac{1}{x}\times x \right\}dx} \right] \\
 & =x\log x-\int{1dx} \\
\end{align}\]
\[=x\log x-x\]
So, \[\int{\log xdx}=x\log x-x\]………………………………….(8)
Now, using equation (8) and on simplifying equation (5), we get
\[=\log x\left( x\log x-x \right)-\left[ \int{\left\{ \dfrac{1}{x}\left( x\log x-x \right) \right\}dx} \right]\]
\[=x{{\left( \log x \right)}^{2}}-x\log x-\left[ \int{\log xdx-\int{1dx}} \right]\] ……………………………………..(9)
Now, from equation (8) and equation (9), we get
\[\begin{align}
  & =x{{\left( \log x \right)}^{2}}-x\log x-\left[ x\log x-x-x \right] \\
 & =x{{\left( \log x \right)}^{2}}-x\log x-x\log x+2x \\
 & =x{{\left( \log x \right)}^{2}}-2x\log x+2x \\
\end{align}\]
On taking \[x\] as common in the above equation, we get
\[=x\left[ {{\left( \log x \right)}^{2}}-2\log x+2 \right]\]
Therefore, the value of the expression \[\int{{{\left( \log x \right)}^{2}}dx}\] is equal to \[x\left[ {{\left( \log x \right)}^{2}}-2\log x+2 \right]\] .
Hence, the correct option is (A).


Note:
 For this type of question, apply the integration by parts formula on a prior basis. Using the integration by parts formula can make us realize the integration of which function is required. For instance, here in this question on applying integration by parts formula for the expression \[\int{{{\left( \log x \right)}^{2}}dx}\] , the integration of the function \[\int{\log xdx}\] is required.