Question

How do you integrate \[\int {\dfrac{{{{(\ln x)}^2}}}{x}} \] by integration by parts method?

Answer 1

Hint: According to the integration by parts method, the integration of the product of two functions is equal to the difference of the product of the first function and the integration of the second function and integration of the product of the derivative of the first function and the integration of the second function, that is, \[\int {u(x)v(x) = u(x)\int {v(x)} } - \int {u'(x)(\int {v(x} )} \] . So for solving the given question by integration by parts method, we have to first express \[\dfrac{{{{(\ln x)}^2}}}{x}\] as the product of two functions and then solve using the above-mentioned definition.

Complete step by step solution:
We can write $\dfrac{{{{(\ln x)}^2}}}{x}$ as the product of ${(\ln x)^2}$ and $\dfrac{1}{x}$ .
$ \Rightarrow \int {{{(\ln x)}^2}.\dfrac{1}{x}dx} $
Applying integration by parts method –
 \[
  \int {\dfrac{{{{(\ln x)}^2}}}{x}} dx = {(\ln x)^2}\int {\dfrac{1}{x}dx} - \int {\dfrac{{d{{(\ln x)}^2}}}{{dx}}(\int {\dfrac{1}{x}dx} )dx} \\
   \Rightarrow \int {\dfrac{{{{(\ln x)}^2}}}{x}} dx = {(\ln x)^2}\ln x - \int {2\ln x.\dfrac{1}{x}.\ln x} dx + c \\
   \Rightarrow \int {\dfrac{{{{(\ln x)}^2}}}{x}} dx = {(\ln x)^3} - 2\int {\dfrac{{{{(\ln x)}^2}}}{x}} dx + c \;
 \]
Now taking \[2\int {\dfrac{{{{(\ln x)}^2}}}{x}} dx\] to the left-hand side, we get –
$
  3\int {\dfrac{{{{(\ln x)}^2}}}{x}} dx = {(\ln x)^3} + c \\
   \Rightarrow \int {\dfrac{{{{(\ln x)}^2}}}{x}} dx = \dfrac{1}{3}{(\ln x)^3} + c \;
 $
Hence \[\int {\dfrac{{{{(\ln x)}^2}}}{x}} \] is equal to $\dfrac{1}{3}{(\ln x)^3} + c$ .
So, the correct answer is “$\dfrac{1}{3}{(\ln x)^3} + c$”.

Note: n integration, we are the differentiation of a function and we have to find the function. A function can have an infinite number of integrals as we can get different values of integral by varying the value of the arbitrary constant but every function has a unique derivative. For solving such types of questions, we must know the integration of basic functions like $\int {\ln x} ,\,\int {\dfrac{1}{x},\,etc.} $ While applying the integration by parts method, one might get confused about which function to take as the first function and which function as the second, to resolve this confusion, we have a rule called ILATE (I: Inverse trigonometric function, L: Logarithm function, A: Algebraic function, T: Trigonometric function, E: Exponential function). This rule tells us the order of preference for different types of functions.