Integral Of 1 X 1 X 2

Decoding the Integral: A Deep Dive into ∫1/(x(x+2)) dx

The seemingly simple integral, ∫1/(x(x+2)) dx, presents a wonderful opportunity to explore several crucial concepts in calculus. While the integrand might look intimidating at first glance, employing the right techniques reveals a straightforward solution. This article will not only guide you through the solution but also delve into the underlying principles, providing a comprehensive understanding of the process. We'll explore partial fraction decomposition, a powerful tool in integral calculus, and discuss its broader applications.

Understanding the Problem: Partial Fraction Decomposition

The integral ∫1/(x(x+2)) dx cannot be solved directly using standard integration rules. The presence of the product of two linear terms in the denominator necessitates a technique called partial fraction decomposition. This method allows us to break down a complex rational function (a fraction where the numerator and denominator are polynomials) into simpler fractions that are easier to integrate.

The fundamental principle behind partial fraction decomposition is to express the rational function as a sum of simpler fractions, each with a denominator that is a factor of the original denominator. In our case, the denominator x(x+2) has two distinct linear factors: x and (x+2). Therefore, we can write:

1/(x(x+2)) = A/x + B/(x+2)

where A and B are constants that we need to determine.

Solving for the Constants: A and B

To find the values of A and B, we need to eliminate the denominators. We do this by multiplying both sides of the equation by x(x+2):

1 = A(x+2) + Bx

This equation must hold true for all values of x. We can strategically choose values of x that simplify the equation and allow us to solve for A and B.

Method 1: Equating Coefficients

We can expand the right side of the equation:

1 = Ax + 2A + Bx

1 = (A+B)x + 2A

Now, we equate the coefficients of the corresponding powers of x:

• Coefficient of x: A + B = 0
• Constant term: 2A = 1

From the second equation, we find A = 1/2. Substituting this into the first equation gives B = -1/2.

Method 2: Substituting Values of x

Alternatively, we can choose specific values of x to simplify the equation.

• Let x = 0: 1 = A(0+2) + B(0) => 1 = 2A => A = 1/2
• Let x = -2: 1 = A(-2+2) + B(-2) => 1 = -2B => B = -1/2

Both methods yield the same result: A = 1/2 and B = -1/2.

Integrating the Partial Fractions

Now that we have the values of A and B, we can rewrite the original integral as:

∫1/(x(x+2)) dx = ∫(1/2)/x dx + ∫(-1/2)/(x+2) dx

These integrals are straightforward to solve using the basic rule for integrating 1/u du = ln|u| + C, where C is the constant of integration:

∫(1/2)/x dx = (1/2)ln|x| + C₁

∫(-1/2)/(x+2) dx = (-1/2)ln|x+2| + C₂

Combining these results, we obtain the final solution:

∫1/(x(x+2)) dx = (1/2)ln|x| - (1/2)ln|x+2| + C

where C = C₁ + C₂ is the overall constant of integration.

Simplifying the Solution: Logarithmic Properties

We can further simplify the solution using the logarithmic property ln(a) - ln(b) = ln(a/b):

∫1/(x(x+2)) dx = (1/2)ln|x/(x+2)| + C

This is the most concise form of the solution.

Applications of Partial Fraction Decomposition

Partial fraction decomposition is not just a technique for solving specific integrals; it's a fundamental tool with far-reaching applications in various areas of mathematics and engineering. Here are some examples:

• Calculus: Solving integrals involving rational functions with higher-degree polynomials in the denominator. The complexity increases with the degree of the polynomials, often requiring more sophisticated techniques to factor the denominator and solve for the constants.

• Differential Equations: Solving certain types of differential equations. The method appears when dealing with rational functions in the context of solving for solutions.

• Signal Processing: Analyzing and designing systems involving linear time-invariant (LTI) systems. Partial fraction decomposition facilitates breaking down complex system responses into simpler components for easier analysis.

• Control Systems: Analyzing and designing control systems, often involving transfer functions that are rational functions. Decomposition helps understand the system's response characteristics.

• Probability and Statistics: In probability theory, partial fraction decomposition can be helpful when working with probability distributions or generating functions.

Extending the Concept: More Complex Denominators

The example we've explored involves a denominator with distinct linear factors. However, partial fraction decomposition can also handle more complex scenarios:

• Repeated Linear Factors: If the denominator has repeated linear factors (e.g., (x-a)², (x-a)³), the partial fraction decomposition includes terms with increasing powers of the repeated factor in the denominator.

• Irreducible Quadratic Factors: If the denominator contains irreducible quadratic factors (quadratic expressions that cannot be factored into real linear factors), the corresponding terms in the partial fraction decomposition have a linear numerator over the quadratic factor.

These more complex cases require more intricate algebraic manipulations to determine the constants, but the underlying principle remains the same: break down the complex fraction into simpler, integrable parts.

Conclusion: Mastering a Powerful Tool

The seemingly straightforward integral ∫1/(x(x+2)) dx serves as an excellent illustration of the power and versatility of partial fraction decomposition. This technique is an essential skill for anyone working with calculus and its applications. Understanding the method and its various extensions equips you with a robust tool for tackling more complex problems in diverse fields. By mastering partial fraction decomposition, you unlock a significant door in understanding advanced mathematical concepts and their practical uses. Remember to always check your solution and simplify where possible, leveraging the power of logarithmic properties to achieve the most elegant and efficient results. Practice is key – the more you work with this technique, the more intuitive and comfortable you'll become with its application.