Problem 20 - Calculus Early Transcendentals

Question

Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{1} \frac{x \ln ^{4}\left(x^{2}+1\right)}{x^{2}+1} d x$$

Step-by-Step Solution

Answer

The value of the definite integral is: $$\frac{(\ln(2))^5}{10}$$.

Step 1: Substitution

Let's make a substitution that will simplify the function inside the integral. We can set \(u = \ln(x^{2}+1)\). Then, \(du = \frac{2x}{x^2 + 1}dx\). We can multiply both sides by \(\frac12\) to have \(x\) only in one side and obtain, \(\frac12 du = \frac{x}{x^2 + 1}dx\). Since we are dealing with definite integration, we need to change the limits of integration as well. Let's plug in the limits of integration into the substitution: For \(x=0\), we have \(u=\ln(0^2+1)=\ln(1)=0\). For \(x=1\), we have \(u=\ln(1^2+1)=\ln(2)\). Now we can rewrite the integral: $$\int_{0}^{1} \frac{x \ln ^{4}\left(x^{2}+1\right)}{x^{2}+1} d x =\int_{0}^{\ln(2)} \frac{1}{2}u^{4}du$$

Step 2: Integrate

Now, we need to integrate the function inside the integral with respect to \(u\): $$\int_{0}^{\ln(2)} \frac{1}{2}u^{4}du = \frac{1}{2}\int_{0}^{\ln(2)} u^{4} du$$ Integrating \(u^4\) with respect to \(u\), we get: \(\frac{1}{2}\int u^{4} du = \frac{1}{2} \cdot \frac{1}{5} u^5 = \frac{1}{10}u^5\) Now, we need to evaluate the antiderivative at the new limits of integration. After evaluating, we'll have: $$\frac{1}{10}u^5\Big|_{0}^{\ln(2)}=\frac{1}{10}\left((\ln(2))^5 - (0)^5\right)$$

Step 3: Final Result

Finally, we simplify the expression: $$\frac{1}{10}\left((\ln(2))^5 - (0)^5\right)=\frac{(\ln(2))^5}{10}$$ Thus, the value of the definite integral is: $$\int_{0}^{1} \frac{x \ln ^{4}\left(x^{2}+1\right)}{x^{2}+1} d x = \frac{(\ln(2))^5}{10}$$

Key Concepts

Substitution Method

The substitution method is a technique used to simplify an integral by changing variables. It helps transform a complex equation into a simpler form, making integration manageable. In the given problem, we have employed this method by substituting the expression inside the integral with a new variable. Here's how it works:

• We identify a function within the integral, say \( f(x) \), that forms a part of a more complex term.
• We substitute \( u = f(x) \), simplifying the integrand.
• Next, we find \( du \), the differential of \( u \), in terms of \( dx \) and manipulate it to help us integrate more easily.

We often need to replace the limits of integration because of the variable change. By calculating \( u \) at the original limits, the new integral will only involve \( u \), simplifying it significantly for further computation.

Definite Integrals

Definite integrals extend the concept of finding the antiderivative of a function over a specific interval. In a definite integral, the integral's limits, indicated by the numbers above and below the integral sign, represent the bounds within which we calculate the area under a curve.

• The process includes evaluating the antiderivative at the upper limit and subtracting the value at the lower limit.
• This provides the total "signed" area under the function's curve from one limit to the other, considering any areas below the x-axis as negative.
• Definite integrals yield a numerical value, whereas indefinite integrals produce a function.

In this exercise, after using the substitution method, we changed the limits of integration accordingly from \( x \) to \( u \), making it easier to integrate over the new limits of \( \ln(1) \) to \( \ln(2) \).

Logarithmic Functions

Logarithmic functions are important in calculus, especially when dealing with integration techniques. A logarithmic function typically takes a form like \( \ln(x) \), where "ln" denotes the natural logarithm. The properties of logarithms can simplify many integration problems.

• Logarithmic functions help convert multiplication into addition, which can simplify complex expressions.
• They have well-known rules, such as \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(a^b) = b\ln(a) \).
• Understanding these properties is critical as they often lead to simplification in problems involving exponential growth or decay.

In the original problem, the function \( \ln(x^2 + 1) \) appears inside the integral. By acknowledging its form and applying a suitable substitution, we transformed this complex term, enabling us to simplify and solve the integral.