Question

How do you evaluate the integral of \[{\left( {\ln x} \right)^2}dx\]?

Answer 1

Hint: In the given question, we have been given a function. The function is a logarithmic function. This logarithmic function contains a variable as an argument. Then this logarithmic function is raised to a power. We have to calculate the value of the integral of this whole logarithmic function. To solve the integral, we are going to need to use integration by parts.

Formula Used:
In the question, we are going to use the formula of integration by parts:
\[\int {udv} = uv - \int {vdu} \]

Complete step-by-step answer:
The given expression to be integrated is \[{\left( {\ln x} \right)^2}dx\].
Let \[u = {\left( {\ln x} \right)^2}\] and \[dv = dx\]
Then, we can say that,
\[du = \dfrac{{2\ln x}}{x}\] and also, \[v = x\]
Using the formula of integration by parts, we get:
$\Rightarrow$ \[\int {{{\left( {\ln x} \right)}^2}dx = x{{\left( {\ln x} \right)}^2} - 2\int {\ln xdx} } \]
Now, we know that
$\Rightarrow$ \[\int {\ln x dx = x\ln x - x + c} \]
 Hence by replacing the values we get,
$\Rightarrow$ \[\int {{{\left( {\ln x} \right)}^2}dx = x{{\left( {\ln x} \right)}^2} - 2\left( {x\ln x - x} \right) + c} \]
Therefore, on simplifying we get, \[\int {{{\left( {\ln x} \right)}^2}dx = x{{\left( {\ln x} \right)}^2} - 2x\ln x + 2x + c} \].

Note: In the given question, we had to find the value of the integral of a logarithmic function. There can be many ways by which we can attempt to find the answer, but the only correct way is to solve by using integration by parts. We figure that out by seeing the \[dx\] part – the differential is of the argument of the determining function – the logarithmic function. So, we must know the concept of the things being used.