Problem 42 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } , \text { evaluate each definite integral. } $$ $$ \int_{2}^{3} \frac{1}{1-x^{2}} d x $$

Step-by-Step Solution

Answer

The integral is undefined in the real number domain.

Step 1: Identify the Integral Type

The given integral is \( \int_{2}^{3} \frac{1}{1-x^{2}} \, dx \). This is a definite integral, and the integrand is a rational function which resembles the form of an inverse hyperbolic function.

Step 2: Recognize the Antiderivative

Notice that \( \frac{1}{1-x^2} \) can be transformed using the formula for the antiderivative of the inverse hyperbolic tangent function. Specifically, \( \frac{1}{1-x^2} \) is the derivative of \( \tanh^{-1}(x) \) (inverse hyperbolic tangent).

Step 3: Calculate the Antiderivative

Given that \( \frac{d}{dx} (\tanh^{-1}(x)) = \frac{1}{1-x^2} \), we identify the antiderivative as \( \tanh^{-1}(x) \).

Step 4: Evaluate the Definite Integral

Substitute the upper and lower limits into the antiderivative: \[ \int_{2}^{3} \frac{1}{1-x^2} \, dx = \left[ \tanh^{-1}(x) \right]_2^3 = \tanh^{-1}(3) - \tanh^{-1}(2) \].

Step 5: Consider Real Values

The function \( \tanh^{-1}(x) \) is only real for \(|x| < 1\). Since 2 and 3 are outside this domain, the integral evaluates to complex numbers or is considered undefined in the realm of real numbers.

Step 6: Conclude the Evaluation

Since \( \tanh^{-1}(x) \) is undefined for values outside \(|x| < 1\) in the real domain, the integral cannot be evaluated with real numbers. Hence, the integral is undefined over the interval [2, 3].

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of trigonometric functions but for hyperbolas instead of circles. In the case of hyperbolic functions, the inverse hyperbolic tangent, written as \( \tanh^{-1}(x) \), is especially relevant. It's important to understand the domain constraints of these functions.
The function \( \tanh^{-1}(x) \) is defined only for \( |x| < 1 \) in the real number system. This restriction arises because as \( x \) approaches ±1, \( \tanh^{-1}(x) \) approaches infinity. Thus, it cannot accept real numbers outside the interval \((-1, 1)\).
You might wonder why we would use a function with such restrictions. The appeal of inverse hyperbolic functions is their close relationships to logarithmic functions, making them useful for solving integrals related to rational functions or those that transform into these integrals.

Rational Functions

Rational functions are expressions that are formed by the division of two polynomials. For example, \( \frac{1}{1-x^2} \) is a classic rational function, where the numerator is a constant, and the denominator is a quadratic polynomial.
In calculus, when you encounter a rational function inside an integral, it typically calls for specific strategies. Some rational functions are integrable using elementary methods, while others require substitution or recognition of certain derivatives.

• Substitution involves changing variables to simplify the integral.
• Recognition uses known antiderivatives, like those involving inverse hyperbolic functions, to solve the integral directly.

Understanding the properties and peculiarities of rational functions is crucial because they often appear in real-life applications like physics and engineering, especially in expressions that model growth, decay, or motion.

Antiderivative

Antiderivatives are functions that represent the "inverse" of differentiation. When you find an antiderivative, you're essentially searching for a function that, when differentiated, returns the original function you have.
In the context of our integral \( \int \frac{1}{1-x^2} \, dx \), recognizing that this is related to the derivative of \( \tanh^{-1}(x) \) is key. This recognition allows you to take a shortcut to solving integrals by leveraging known antiderivatives.
Given the significance of the antiderivative, it is vital to know the common antiderivatives and identities:

• \( \frac{d}{dx} (\tanh^{-1}(x)) = \frac{1}{1-x^2} \) is one such identity, indicating the relation between \( \tanh^{-1}(x) \) and our function.
• Antiderivatives help simplify the process of evaluating definite integrals by providing a function to evaluate at the bounds.

Recognizing antiderivatives can simplify complex calculus problems, turning seemingly challenging integrals into straightforward applications of known identities.