Question

Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms. $$\int_{1}^{e^{2}} \frac{d x}{x \sqrt{\ln ^{2} x+1}}$$

Short answer

Question: Evaluate the definite integral $$\int_{1}^{e^{2}} \frac{d x}{x \sqrt{\ln ^{2} x+1}}$$ using the relationship between natural logarithm function and integral. Answer: $$\int_{1}^{e^{2}} \frac{d x}{x \sqrt{\ln ^{2} x+1}} = \ln(2+\sqrt{5})$$

Step-by-step solution

Identify the Substitution

To express this integral in terms of logarithm, let's set \(u=\ln{x}\). This substitution will transform the integral into a more tractable form.

Find \(du\) and \(dx\)

Now, differentiate both sides of the equation obtained in Step 1: $$\frac{d u}{d x}=\frac{d}{d x}(\ln x).$$ Using the chain rule, we have $$\frac{d u}{d x}=\frac{1}{x}.$$ Multiplying both sides by \(dx\), we get: $$du=\frac{1}{x}dx.$$

Change the Limits of Integration

Now, we will change the limits of integration according to our substitution: \(u\) when \(x=1\): \(u=\ln{1}=0\) \(u\) when \(x=e^2\): \(u=\ln{e^2}=2\)

Substitute \(u\), \(du\), and the Limits of Integration

Now substitute these new limits of integration and the expressions for \(dx\) and \(u\) into the integral: $$\int_{1}^{e^{2}} \frac{d x}{x \sqrt{\ln ^{2} x+1}} = \int_0^2\frac{1}{\sqrt{u^2+1}} \ du$$

Evaluate the Integral

Now our integral has the form of $$\int\frac{du}{\sqrt{u^2+1}}.$$ The antiderivative of this function is \(\sinh^{-1}(u)=\ln(u+\sqrt{u^2+1})+C\). So our definite integral is: $$\int_0^2\frac{1}{\sqrt{u^2+1}} \ du = \left[\ln(u+\sqrt{u^2+1})\right]_0^2$$

Substitute Back the Variable

Now substitute \(u=\ln{x}\) to get the final result: $$\left[\ln(\ln(x)+\sqrt{\ln^2(x)+1})\right]_1^{e^2}$$

Evaluate the Definite Integral

Now we compute the result by subtracting the values of the function at the upper and lower limits of integration: $$\left[\ln(\ln(x)+\sqrt{\ln^2(x)+1})\right]_1^{e^2} = \ln(\ln(e^2)+\sqrt{\ln^2(e^2)+1}) - \ln(\ln(1)+\sqrt{\ln^2(1)+1})$$ Simplifying the expression, $$= \ln(2+\sqrt{2^2+1}) - \ln(0+\sqrt{0+1})$$ $$= \ln(2+\sqrt{5}) - \ln(1)$$ $$= \ln\left(\frac{2+\sqrt{5}}{1}\right)$$ Thus, the definite integral evaluates to: $$\int_{1}^{e^{2}} \frac{d x}{x \sqrt{\ln ^{2} x+1}} = \ln(2+\sqrt{5})$$

Key concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify the process of integration. By replacing a part of the integral with a new variable, the integral can often be transformed into a simpler form that is easier to evaluate.

In this specific problem, we used the substitution \( u = \ln{x} \). This substitution is appropriate because it simplifies the complicated expressions involving the natural logarithm into a more manageable form.

Here is a step-by-step breakdown of using the substitution method:

• Identify a part of the integrand (the function being integrated) that can be replaced by a single variable to simplify the integral.
• Differentiate the substitution equation to find \( du \) in terms of \( dx \).
• Substitute \( u \) and \( du \) into the integral, along with new limits if it's definite, which translates our integral into a new variable that is often easier to work with.

This step greatly simplifies integrals especially when dealing with exponents, logarithms, and trigonometric functions.

Limits of Integration

When performing definite integration, it's crucial to adjust the limits of integration if you've made a substitution. These limits define the range over which you're integrating, and changing the variable requires recalibrating this range.

For example, in the given problem, the original limits were \( x = 1 \) to \( x = e^2 \). After substitution with \( u = \ln{x} \), the limits changed to:

• When \( x = 1 \), \( u = \ln(1) = 0 \).
• When \( x = e^2 \), \( u = \ln(e^2) = 2 \).

These new limits of integration \( 0 \) to \( 2 \) are used in the transformed integral. Remember, integration is calculating the area under the curve within specific limits, so if the variable changes, the limits must reflect that transformation too.

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm \( \ln(x) \), are prevalent in calculus. They appear in problems involving exponential growth, decay, and integration where the logarithmic rules come in handy.

In calculus, the derivative of \( \ln(x) \) is \( \frac{1}{x} \), and it frequently plays a role in integration via the substitution method. In the exercise, the use of \( u = \ln{x} \) simplified the integral, as it's easier to work with \( u^2 + 1 \) rather than \( \ln^2(x) + 1 \).

Additionally, expressions involving \( \ln \) often appear in the solutions of integrals, either directly or within other functions. It serves as an essential tool in manipulating expressions to find the antiderivative needed for applying the fundamental theorem of calculus.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as \( \sinh^{-1} \), arise in integration, especially when dealing with expressions of the form \( \sqrt{u^2 + 1} \). These functions, similar to their trigonometric counterparts, have defined antiderivatives.

In our exercise, the integral form \( \int \frac{du}{\sqrt{u^2+1}} \) is directly associated with \( \sinh^{-1}(u) \). This is because the derivative of \( \sinh^{-1}(u) \) is precisely \( \frac{1}{\sqrt{u^2+1}} \), making it a common result when integrating such expressions.

Understanding these functions can vastly simplify complex integrals that may otherwise seem intractable. They broaden the toolkit available for solving integrals, especially when simpler algebraic or trigonometric techniques fall short. As such, they are yet another example of how extensions from basic calculus, like hyperbolic functions, can resolve higher-level problems efficiently.