Question

How do you integrate $\int {{x^2}.{{\left( {\ln x} \right)}^2}} dx$ by integration by parts method?

Answer 1

Hint : Integration by parts is used for integrating the product of two functions. This method is used to solve the integration easily. We know the formula for integrating by parts is given by $\int {u{\text{ }}v{\text{ }}dx = u\int {v{\text{ }}dx - \int {\left( {\dfrac{{du}}{{dx}}\int {v{\text{ }}dx} } \right)} } } dx$ . Since in the given problem we have to produce two functions, we take $u = {\left( {\ln x} \right)^2}$ and $v = {x^2}$ .

Complete step-by-step answer :
Given, $\int {{x^2}.{{\left( {\ln x} \right)}^2}} dx$ .
Let’s take $I = \\\int {{x^2}.{{\left( {\ln x} \right)}^2}} dx$
(because if we keep on integrating it’s never going to end, we need to apply integration by parts infinite times so we took it as ‘I’).
As we know the formula for integration by parts,
$\int {u{\text{ }}v{\text{ }}dx = u\int {v{\text{ }}dx - \int {\left( {\dfrac{{du}}{{dx}}\int {v{\text{ }}dx} } \right)} } } dx$
And we have $u = {\left( {\ln x} \right)^2}$ and $v = {x^2}$ .
Substituting we have
$\int {{x^2}.{{\left( {\ln x} \right)}^2}} dx$
$= {\left( {\ln x} \right)^2}\int {{{\text{x}}^2}{\text{ }}dx - \int {\left( {\dfrac{{d{{\left( {\ln x} \right)}^2}}}{{dx}}\int {{{\text{x}}^2}{\text{ }}dx} } \right)} } dx$
$= \left( {{{\ln }^2}x} \right).\dfrac{{{x^3}}}{3} - \int {\dfrac{{{x^3}}}{3}} .2\ln x.\dfrac{1}{x}dx$
$= \left( {{{\ln }^2}x} \right).\dfrac{{{x^3}}}{3} - \dfrac{2}{3}\int {{x^2}\ln xdx} - - - (1)$
Now we need $\int {{x^2}\ln xdx}$ values we use parts by method to solve it,
here $u = \ln x$ and $v = {x^2}$
$\int {{x^2}\ln xdx} = \ln x\int {{x^2}dx - \int {\dfrac{{d(\ln x)}}{{dx}}\int {{x^2}} } } dx$
$= \ln x.\dfrac{{{x^3}}}{3} - \int {\dfrac{{{x^3}}}{3}.\dfrac{1}{x}dx}$
$= \ln x.\dfrac{{{x^3}}}{3} - \int {\dfrac{{{x^2}}}{3}} dx$
$= \ln x.\dfrac{{{x^3}}}{3} - \dfrac{1}{3}\dfrac{{{x^3}}}{3} + c$
$\int {{x^2}\ln xdx} = \ln x.\dfrac{{{x^3}}}{3} - \dfrac{{{x^3}}}{9} + c$ , where c is the integration constant.
Substituting this in equation (1) we have,
$= \left( {{{\ln }^2}x} \right).\dfrac{{{x^3}}}{3} - \dfrac{2}{3}\left( {\ln x.\dfrac{{{x^3}}}{3} - \dfrac{{{x^3}}}{9}} \right) + C$
Thus we have
$\Rightarrow \int {{x^2}.{{\left( {\ln x} \right)}^2}} dx = \left( {{{\ln }^2}x} \right).\dfrac{{{x^3}}}{3} - \dfrac{2}{3}\left( {\ln x.\dfrac{{{x^3}}}{3} - \dfrac{{{x^3}}}{9}} \right) + C$ , where ‘C’ is the integration constant.
So, the correct answer is “Option C”.

Note : Here we use a rule called ‘ILATE’. The integral of two functions is taken by considering the first term as first function and second term as a second function. This is called the ‘ILATE’ rule. In the given problem we have ${x^2}.{\left( {\ln x} \right)^2}$ . In this we considered ${\left( {\ln x} \right)^2}$ as a first function and ${x^2}$ as a second function. That is the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometry, exponent. As we can see we have trigonometric functions first in the preference and then exponent.