Problem 34 - Calculus: Early Transcendental Functions

Question

One of the two antiderivatives can be determined using basic algebra and the derivative formulas we have presented. Find the antiderivative of this one and label the other \(^{\text {*. }} \mathbf{N} / \mathbf{A}^{\mathbf{\prime}}\) $$\text { (a) } \int\left(\frac{1}{x^{2}}-1\right) d x$$ $$\text { (b) } \int \frac{1}{x^{2}-1} d x$$

Step-by-Step Solution

Answer

The antiderivative of \(\frac{1}{x^{2}}-1\) is \(-\frac{1}{x} - x + C\) and the antiderivative of \(\frac{1}{x^{2}-1}\) is \(\frac{1}{2} ln|x-1| - \frac{1}{2} ln|x+1| + C\).

Step 1: Find the antiderivative of the function in Part a.

Remember that the antiderivative or integral of a function f(x) is another function F(x) whose derivative is f(x). So, to find the antiderivative of \(\frac{1}{x^{2}}-1\), break this expression into two separate integrals, and perform the integration for each: \[\int \frac{1}{x^2} dx - \int 1 dx\] The antiderivative of \(\frac{1}{x^2}\) is \(-\frac{1}{x}\), and the antiderivative of 1 is \(x\). Thus, the antiderivative of the given function is: \(-\frac{1}{x} - x + C\) where \(C\) is the constant of integration.

Step 2: Convert the denominator in Part b to a form suitable for decomposition into partial fractions.

Part b is a bit trickier as it requires a technique called partial fractions decomposition. The expression inside the integral \(\frac{1}{x^2 - 1}\) is a rational function, and it can be decomposed into simpler fractions that are easier to integrate. However, in order to do that, let's first rewrite \(x^2 - 1\) as \((x+1)(x-1)\). Therefore, the function becomes \(\frac{1}{(x+1)(x-1)}\).

Step 3: Decompose into partial fractions and integrate.

Now that the denominator is factored, we can rewrite the expression as two simpler fractions: \(A/(x+1) + B/(x-1)\). Now to find \(A\) and \(B\), one is to set the expression equal to \(\frac{1}{(x+1)(x-1)}\) and solve for \(A\) and \(B\). They are found to be \(A = -1/2\) and \(B = 1/2\). Substitute \(A\) and \(B\) into the original expression to get: \(-\frac{1}{2(x+1)} + \frac{1}{2(x-1)}\). Now integrate this expression to find the antiderivative. The integral of the first part is \(-\frac{1}{2} ln|x+1|\) and for the second part it is \(\frac{1}{2} ln|x-1|\). Thus, the final antiderivative for this function is: \(\frac{1}{2} ln|x-1| - \frac{1}{2} ln|x+1| + C\)

Key Concepts

Antiderivative

In the world of calculus, the antiderivative plays a crucial role. It is essentially the reverse process of taking a derivative. If you think of derivatives as finding the rate of change, antiderivatives find the original function from that rate.
An antiderivative for a function \( f(x) \) is a function \( F(x) \) such that the derivative of \( F(x) \) gives back \( f(x) \). That is, \( F'(x) = f(x) \). Finding an antiderivative is equivalent to solving an indefinite integral:

• The integral \( \int f(x) \, dx \) gives \( F(x) + C \), where \( C \) is the constant of integration.

To solve for the antiderivative:

• Break complex expressions into smaller, easier parts.
• Integrate each component separately using known integral formulas.

For example, consider \( \int \left( \frac{1}{x^2} - 1 \right) \, dx \). This can be split into two parts: \( \int \frac{1}{x^2} \, dx \) and \( \int 1 \, dx \). The antiderivative of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \), and for the constant 1, it is \( x \). So, the overall antiderivative is \( -\frac{1}{x} - x + C \).

Partial Fraction Decomposition

When you have rational functions, which are ratios of polynomials, one effective method for integration is partial fraction decomposition. This technique simplifies complex rational expressions into a sum of simpler fractions, making them easier to integrate.
For instance, consider the integral \( \int \frac{1}{x^2-1} \, dx \). The denominator \( x^2-1 \) can be factored into \( (x+1)(x-1) \). This allows us to decompose the expression into:

• \( \frac{1}{x^2-1} = \frac{A}{x+1} + \frac{B}{x-1} \)

To find the coefficients \( A \) and \( B \), set up an equation:

• \( 1 = A(x-1) + B(x+1) \)

Solve for \( A \) and \( B \) by substituting convenient values or by equating coefficients. For this example, \( A = -\frac{1}{2} \) and \( B = \frac{1}{2} \).
Now, the integral becomes easier to compute as two simpler integrals: \( \int \left(-\frac{1}{2} \right) \ln |x+1| + \int \left( \frac{1}{2} \right) \ln |x-1| \, dx \). This gives the final antiderivative: \( \frac{1}{2} \ln |x-1| - \frac{1}{2} \ln |x+1| + C \).

Rational Functions

Rational functions are expressions of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. These functions can be quite complex, depending on the degree and the nature of the polynomials involved.
They show up often in calculus problems and can be particularly tricky to integrate without additional strategies:

• Use polynomial long division if the degree of \( P(x) \) is higher than \( Q(x) \).
• Apply the method of partial fraction decomposition for a simpler integration process.

In integration, such functions might require extra steps like rewriting the expression to make it compatible for known methods. For example, factoring the denominator as seen in the exercise \( \int \frac{1}{x^2-1} \, dx \).
Once in a decomposed form, each part can be individually integrated using basic integration rules, which simplifies the process considerably.