Problem 47 - Calculus Single Variable

Question

Evaluate each of the integrals. $$ \int\left(\frac{\ln (x)}{x}\right)^{2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{(\ln x)^3}{3} + C \).

Step 1: Identify the integration technique

Recognize that the integral \[ \int \left(\frac{\ln(x)}{x}\right)^2 \, dx \] suggests the use of substitution or integration by parts due to its structure. A common method for integrals involving \(\ln(x)\) is integration by parts.

Step 2: Setup for substitution

Start by making a substitution to simplify the integral. Set \( u = \ln(x) \), which implies \( du = \frac{1}{x} \, dx \). This substitution transforms the integral into an expression in terms of \( u \).

Step 3: Substitute and transform the integral

After substituting \( u = \ln(x) \) and \( du = \frac{1}{x} \, dx \), the integral becomes: \[ \int u^2 \, du \] This is a standard power integral, which can be easily integrated.

Step 4: Integrate the transformed integral

Evaluate the integral \( \int u^2 \, du \):Using the power rule for integration, we get \[ \frac{u^3}{3} + C \] where \( C \) is the constant of integration.

Step 5: Back-substitute to original variable

Now, replace \( u \) with \( \ln(x) \) to express the result in terms of the original variable \( x \):\[ \frac{(\ln(x))^3}{3} + C \] This gives the final solution in terms of \( x \).

Key Concepts

Integration by Parts

Integration by Parts is a powerful technique used to solve integrals where the standard methods don't apply easily. It's particularly useful with products of functions, like polynomials multiplied by logarithms or exponentials. The core principle behind it relies on the product rule for differentiation.The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]Here, you choose one part of the function to differentiate (\( u \)) and another to integrate (\( dv \)). Once you've set these, you:

• Differentiated \( u \) to find \( du \).
• Integrated \( dv \) to find \( v \).
• Substitute into the formula to solve the integral.

In the original exercise, although it was considered as an approach, substitution was a better starting point. Integration by parts may have been more complex given the structure of \( \frac{\ln(x)}{x} \), but recognizing it as a viable option demonstrates its flexibility.Whenever you see an integral involving a product, consider integration by parts. It's a handy tool in your mathematical toolbox.

Substitution Method

The Substitution Method is all about simplifying an integral by changing its variable. This is done by substituting part of the integrand with another variable, typically denoted by \( u \). This transformation should make the integral easier to solve.The general steps are:

• Identify the part of the integrand that could simplify the problem. Substitute it with a new variable \( u \).
• Calculate \( du \) in terms of the original variable and differentials.
• Rewrite the entire integral in terms of \( u \), replacing all instances of the original variable.
• Integrate with respect to \( u \).
• Substitute back the original variable to express the solution in its initial form.

For the integral \( \int \left(\frac{\ln(x)}{x}\right)^2 \, dx \), substitution shines through. Setting \( u = \ln(x) \) makes \( du = \frac{1}{x} \, dx \), leading to the simpler integral \( \int u^2 \, du \). This change converts a complex-looking problem into a straightforward integration task.

Power Rule for Integration

The Power Rule for Integration is a fundamental technique used when integrating polynomial functions. It's quite straightforward to apply and is one of the first tools you learn in calculus for finding antiderivatives.The formula for the Power Rule is:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]Here, \( n \) cannot be \(-1\), since that leads to a logarithmic result instead. In practice, you:

• Identify the power of \( x \) in the expression.
• Add 1 to the power.
• Divide by the new power value for the solution.

The original exercise, after substitution, required the integration of \( u^2 \). Applying the Power Rule here gives us \( \frac{u^3}{3} + C \). This showcases how versatile and essential the rule is when dealing with polynomial terms. The rule simplifies the process of finding antiderivatives, making it invaluable when tackling similar expressions.