Problem 59 - Calculus Early Transcendentals

Question

Definite integrals Evaluate the following definite integrals. Use Theorem \(6.10\) to express your answer in terms of logarithms. $$\int_{1}^{e^{2}} \frac{d x}{x \sqrt{\ln ^{2} x+1}}$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral \(\int_1^{e^2} \frac{dx}{x\sqrt{\ln^2x+1}}\). Answer: \(\ln(2 + \sqrt{5})\).

Step 1: Identify a substitution

The expression inside the square root, \(\ln^2x+1\), suggests we can make the substitution \(u=\ln x\) to simplify the integrand. Now, we need to find the derivative of \(u\) with respect to \(x\).

Step 2: Compute the derivative

Now we compute the derivative of \(u\) with respect to \(x\): $$\frac{du}{dx}=\frac{d(\ln x)}{dx} =\frac{1}{x} \implies du = \frac{1}{x}\,dx\Rightarrow dx=x\,du.$$

Step 3: Transform the integral

Replace \(dx\) and \(\ln x\) in the original integral with their expressions in terms of \(u\): $$\int_1^{e^2} \frac{dx}{x\sqrt{\ln^2x+1}} = \int_{\ln1}^{\ln e^2}{\frac{x\, du}{x\sqrt{u^2+1}}} = \int_{0}^{2} \frac{du}{\sqrt{u^2+1}}.$$

Step 4: Evaluate the transformed integral

Now, we can integrate the transformed function: $$\int \frac{du}{\sqrt{u^2 + 1}} = \ln|u + \sqrt{u^2 + 1}| + C.$$

Step 5: Substitute back

Replace \(u\) with its expression in terms of \(x\): $$\ln|u + \sqrt{u^2 + 1}| + C = \ln|\ln x + \sqrt{(\ln x)^2 + 1}| + C.$$

Step 6: Apply the Fundamental Theorem of Calculus

Now we are ready to apply the Fundamental Theorem of Calculus to find the definite integral: $$\int_1^{e^2} \frac{dx}{x\sqrt{\ln^2x+1}} = \left[\ln|\ln x + \sqrt{(\ln x)^2 + 1}|\right]_1^{e^2}.$$

Step 7: Calculate the result

Finally, we can evaluate the antiderivative at the two endpoints and subtract their values to find the definite integral: \begin{align*} \int_1^{e^2} \frac{dx}{x\sqrt{\ln^2x+1}} &= \left[\ln|\ln x + \sqrt{(\ln x)^2 + 1}|\right]_1^{e^2} \\ &= \ln|\ln e^2 + \sqrt{(\ln e^2)^2 + 1}| - \ln|\ln 1 + \sqrt{(\ln 1)^2 + 1}| \\ &= \ln|2 + \sqrt{2^2 + 1}| - \ln|0 + \sqrt{0 + 1}| \\ &= \ln(2 + \sqrt{5}) - \ln(1) \\ &= \ln(2 + \sqrt{5}). \end{align*} Thus, the value of the definite integral is \(\ln(2 + \sqrt{5})\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify complex integrals. In this exercise, we identified the expression \( \ln^2x + 1 \) inside the square root as a good candidate for substitution. Here's how it works:

You pick a substitution variable, say \( u \), to simplify the integral.

• For our problem, we chose \( u = \ln x \).
• After differentiating, we found that \( \frac{du}{dx} = \frac{1}{x} \), which rearranges to \( du = \frac{1}{x}dx \).

We then replace all instances of \( x \) and \( dx \) in the original integral with expressions involving \( u \) and \( du \). This turns our integral into an easier form that is much simpler to solve. This substitution reduced our integral to:\[ \int_{0}^{2} \frac{du}{\sqrt{u^2+1}} \] a straightforward expression to work with.

This method often turns complex tasks into simpler ones by transforming variables into something more straightforward. Remember, choosing the right \( u \) is key to simplifying the integrals efficiently.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus plays a crucial role in evaluating definite integrals. It connects differentiation and integration, the two main operations in calculus.

In this exercise, after substituting back and finding the antiderivative:\[ \ln|u + \sqrt{u^2 + 1}| + C \] we applied this theorem to calculate the exact value of the definite integral. Here's a quick reminder of how the theorem works:

• It states that if you have an antiderivative \( F \) of a function \( f \), then the definite integral of \( f \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).

Using this theorem, we evaluated the antiderivative at the boundaries \( 1 \) and \( e^2 \), subtracting the two results. This step is what ultimately gave us the final expression of \( \ln(2 + \sqrt{5}) \).

The fundamental theorem not only puts a concrete number to the integral but also reaffirms the relationship between differentiation and integration.

Transformation of Integrals

Transforming integrals is a method that takes our calculus work from complex to manageable. It's about changing variables and boundaries to simplify solving.

In this exercise, transformation played a significant role when changing the integral bounds along with the substitution:

• We initially had bounds from \( x = 1 \) to \( x = e^2 \).
• After choosing the substitution \( u = \ln x \), these bounds transformed to \( u = 0 \) to \( u = 2 \).

Adjusting limits is an essential part of substitution, allowing the integral to remain complete and correct.

Sometimes, students forget this part during substitution, leading to incorrect solutions. Transforming both bounds and integrand helps focus the work on simpler mathematical expressions, like reducing our integral to: \( \int_{0}^{2} \frac{du}{\sqrt{u^2+1}} \).

This step ensures accuracy in your integration process and makes it easier to visualize and compute results.