Question

Evaluate the indefinite integral, \[\int {x{{(lnx)}^2}dx} \].

Answer 1

Hint: There are various ways that are used to solve indefinite integrals. Integration by parts is one of methods to solve such questions as given. It is otherwise referred to as, the product rule of integration. Recall the formula, put the values and arrive at the solution by solving.

Complete step by step answer:
Integration by parts as the name suggests is basically the breaking up of the integral in parts and then integrating the same. The formula for this is, \[\int {udv} = uv - \int {vdu} \]. Here, “u” is the 1st function and “du” is its differentiation. “dv” is another function and “v” is its integration.
Let \[u = {(ln(x))^2}\,\& dv = xdx\].
We need to find \[du\]and \[v\] from the above 2 assumptions. That will give us,
\[du = {{2ln(x)}}{{xdx}}\,\& v = \dfrac{{{x^2}}}{2}\]
So, putting all these values in their appropriate places in the formula for integration by parts, the resultant will be,
\[\int {x{{(ln(x))}^2}dx} = \dfrac{{{{(xln(x))}^2}}}{2} - \int {xln(x)dx} \]
Let us name the 1st part of the RHS as part-I and the 2nd part as part-II.
We see that again we have to apply integration by parts on part-II. This is called recursive integration by parts.
Integration by parts for the second integral:
Let \[du = \dfrac{1}{x}dx\& v = \dfrac{{{x^2}}}{2}\]\[u = ln(x)\& dv = xdx\;\].
Now proceeding the same way as above,
\[\int {xln(x)dx} = \dfrac{{{x^2}}}{2}ln(x) - \dfrac{1}{2}\int {xdx} \]\[\]
Putting the value of part-II in the main question, we have
\[\int {x{{(ln(x))}^2}dx} = \dfrac{{{{(xln(x))}^2}}}{2} - \dfrac{{{x^2}}}{2}ln(x) + \dfrac{1}{2}\int {xdx} \]
\[\therefore\int {x{{(ln(x))}^2}dx} = \dfrac{{{{(xln(x))}^2}}}{2} - \dfrac{{{x^2}}}{2}ln(x) + \dfrac{1}{4}{x^2} + C\]

Note:An indefinite integral is one function that takes the antiderivative of another function. It's visually represented as an integral symbol, a function, then a dx at the tip. The indefinite integral may have different answers based on the way it is solved but definite integrals have one and only one answer.