Problem 418 - Calculus Volume 1

Question

For the following exercises, find the antiderivatives for the functions. $$\int-\frac{2 x}{x^{4}-1}$$

Step-by-Step Solution

Answer

\( \frac{1}{2} \ln \left| \frac{x^2-1}{x^2+1} \right| + C \)

Step 1: Analyze the Integrand

The function to integrate is given as \(-\frac{2x}{x^4-1}\). First, observe that \(x^4 - 1\) can be factored using the difference of squares: \((x^2-1)(x^2+1)\). This suggests we might use partial fraction decomposition.

Step 2: Set up Partial Fraction Decomposition

Decompose \(\frac{2x}{(x^2-1)(x^2+1)}\) into partial fractions. Assume \(\frac{2x}{(x^2-1)(x^2+1)} = \frac{Ax+B}{x^2-1} + \frac{Cx+D}{x^2+1}\). We need to determine the constants \(A\), \(B\), \(C\), and \(D\).

Step 3: Find Partial Fraction Coefficients

Multiply through by \((x^2-1)(x^2+1)\) to clear the denominators and equate coefficients given the polynomial: \[2x = (Ax+B)(x^2+1) + (Cx+D)(x^2-1)\]. Expanding and equating coefficients, solve for \(A\), \(B\), \(C\), and \(D\).

Step 4: Perform Algebraic Simplification

After simplification from expanding, solve: \[2x = Ax^3 + Ax + Bx^2 + B + Cx^3 - Cx + Dx^2 - D\]. By comparing coefficients of like terms, determine the values. Solve this system of equations: \(A+C=0\), \(B+D=0\), \(A-C=2\), and \(B-D=0\).

Step 5: Solve the System of Equations

From \(A+C=0\), we get \(C=-A\). From \(A-C=2\), substituting \(C=-A\), this yields \(A+C=0\), then \(A+A=2\) results \(A=1\). Hence, \(C=-1\). From \(B+D=0\) and \(B-D=0\), we get both \(B=0\) and \(D=0\).

Step 6: Integrate Each Partial Fraction

With the values found, the decomposition \(\frac{2x}{(x^2-1)(x^2+1)} = \frac{x}{x^2-1} - \frac{x}{x^2+1}\). Now integrate each: for \(\frac{x}{x^2-1}\) perform substitution \(u = x^2 - 1\), and \(du = 2xdx\); for \(\frac{x}{x^2+1}\) use \(u = x^2 + 1\), and \(du = 2xdx\).

Step 7: Compute the Integrals

Integrating, we find: \[\int \frac{x}{x^2-1} dx = \frac{1}{2} \ln|x^2-1| + C_1\] and \[\int \frac{x}{x^2+1} dx = \frac{1}{2} \ln|x^2+1| + C_2\].

Step 8: Combine the Results

The antiderivative is \( \frac{1}{2} \ln|x^2-1| - \frac{1}{2} \ln|x^2+1| + C\). Thus, the solution in compact form is: \[ \frac{1}{2} \ln \left| \frac{x^2-1}{x^2+1} \right| + C \].

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a powerful method used to simplify the integration of rational functions. When faced with a fraction such as \(\frac{2x}{x^4-1}\), our goal is to break it down into simpler, more manageable parts. This process often involves the following steps:

• Factoring the denominator if possible. In this exercise, \(x^4 - 1\) is factored into \( (x^2-1)(x^2+1) \) using the difference of squares.
• Setting up the decomposition by assuming the original fraction can be expressed as a sum of simpler fractions. Here, it's presumed \( \frac{Ax+B}{x^2-1} + \frac{Cx+D}{x^2+1} \).
• Solving for unknown coefficients \(A, B, C,\) and \(D\) by equating coefficients from both sides after multiplying by the common denominator to eliminate fractions.

Once these coefficients are determined, integration becomes far simpler, as you'll integrate each simple fraction separately.

Integration Techniques

Integration techniques are essential tools in calculus that allow us to compute antiderivatives of functions. In the given solution, after partial fraction decomposition, integration becomes straightforward. Here are some tips and common techniques utilized:

• Substitution: This technique involves changing the variable of integration to simplify the integrand. For instance, in this exercise, \(u = x^2 - 1\) and \(u = x^2 + 1\) were used, which simplifies integration to a natural logarithm form.
• Recognition: When a derivative form is recognized, like \(\frac{du}{u}\), it directly integrates into \ln|u|\. Hence, the integral of our simplified fractions becomes a matter of recognizing these patterns.

Practicing different techniques helps in efficiently solving various integration problems. Understanding when to apply substitution or recognize derivative forms is key.

Differential Equations

Differential equations are equations involving derivatives, and often require solving to find a function. Though not directly part of this exercise, understanding antiderivatives and integration techniques is crucial when working with differential equations.

• Solutions typically involve finding the antiderivative and applying initial conditions to solve for constants if applicable.
• Separation of Variables is one technique that often appears in solving differential equations. Similar to partial fraction decomposition, we rearrange the equation to integrate each side independently.
• Antiderivatives are fundamental when reconstructing the original function from its derivative, which is often the goal in solving a differential equation.

Mastering the process of finding antiderivatives is an essential skill in broader applications like modeling natural phenomena, where differential equations frequently appear.