Question

Evaluate the following integrals. $$\int_{1}^{e^{2}} \frac{\ln ^{2}\left(x^{2}\right)}{x} d x$$

Short answer

Question: Evaluate the integral \(\int_{1}^{e^2} \frac{\ln^2(x^2)}{x} dx\). Answer: \(\frac{32}{3}\)

Step-by-step solution

Simplify the expression \(\ln^2(x^2)\) using logarithm properties

Using the logarithm properties, we can simplify the expression inside the integral as follows: $$\ln^2(x^2) = (\ln(x^2))^2 = (2\ln(x))^2 = 4\ln^2(x)$$ So, the integral becomes: $$\int_{1}^{e^2} \frac{4\ln^2(x)}{x} dx$$

Apply the power rule for integration

To integrate the expression, we first multiply 4 outside of the integral and then apply the power rule for integration: $$4\int_{1}^{e^2} \frac{\ln^2(x)}{x} dx = 4\int_{1}^{e^2} \frac{\ln^2(x)}{x} dx$$ Let \(u = \ln(x)\), then \(\frac{du}{dx} = \frac{1}{x}\), and \(dx = x\, du\): $$4\int_{1}^{e^2} \frac{\ln^2(x)}{x} dx = 4\int_{0}^{2} u^2 du$$ Now, we apply the power rule for integration: $$4\int_{0}^{2} u^2 du = 4\left[\frac{u^3}{3}\right]_0^2$$ Evaluate the integral by plugging in the limits: $$4\left[\frac{u^3}{3}\right]_0^2 = 4\left(\frac{2^3}{3} - \frac{0^3}{3}\right) = 4\left(\frac{8}{3}\right) = \frac{32}{3}$$ So, the value of the integral is: $$\int_{1}^{e^2} \frac{\ln^2(x^2)}{x} dx = \frac{32}{3}$$

Key concepts

Integration Techniques

Integration is a fundamental technique in calculus, often used to find areas, volumes, and central points among other things. In our example, to solve the definite integral \( \int_{1}^{e^{2}} \frac{\ln ^{2}\left(x^{2}\right)}{x} dx \), we must employ various integration techniques that make the process manageable. The first technique involves simplifying the integrand using properties of logarithms. Following this, we use substitution to transform the integrand into a more familiar form. This step is crucial as it converts a seemingly complicated integral into a basic polynomial, which can then be handled with the power rule. Substitution is often used when an integrand contains a function and its derivative. Here, after setting \( u = \ln(x) \), the derivative of \( u \) with respect to \( x \) appears in the integrand, making it perfect for this method. After substituting, we can apply the power rule seamlessly. These strategic moves showcase the importance of mastering various integration techniques to tackle a wide range of problems.

Logarithm Properties

Understanding logarithm properties is crucial in simplifying complex expressions before integrating. In the given exercise, we use the property \( \ln(a^b) = b \ln(a) \) to transform \( \ln^2(x^2) \) into \(4\ln^2(x)\). This fundamental property tells us that we can bring the exponent down in front of the log, effectively simplifying the expression. Other logarithm properties that can be handy include the product rule \( \ln(xy) = \ln(x) + \ln(y) \) and the quotient rule \( \ln(x/y) = \ln(x) - \ln(y) \). By incorporating these properties strategically, we can often reduce integrals to forms that are much easier to work with. This is essential in calculus, as the initial form of an integrand might not be readily integrable without algebraic manipulation. Without these algebraic tools, we could easily find ourselves stuck or working much harder than necessary to solve an integral.

Power Rule for Integration

The power rule for integration is a fundamental concept to master when learning calculus. In its simplest term, the power rule states that the integral of \(x^n\) with respect to \(x\) is \(\frac{1}{n+1}x^{n+1}\), provided \(neq -1\). After applying the substitution \(u = \ln(x)\), we observe that our integral \(4\int_{0}^{2} u^2 du\) presents a situation where the power rule can be directly applied. The outcome, \(\frac{1}{3}u^{3}\), is then evaluated from the bounds 0 to 2. By taking the difference between the upper and lower bounds, we calculate the definite integral's value. Remember, though, when using the power rule to integrate functions like \(\frac{ln^2(x)}{x}\), a substitution is often necessary first to get the integrand into the form \(x^n\), so the power rule can be applied. Mastery of the power rule enables students to quickly and accurately integrate polynomials, a common requirement in many calculus problems.