Solving The Integral Of (2-x)/(1-x) Dx A Step-by-Step Guide

Integrating rational functions is a fundamental skill in calculus, and this article will provide a comprehensive guide to solving the integral ∫(2-x)/(1-x) dx. We'll break down the process step-by-step, ensuring a clear understanding of the techniques involved. This exploration will not only equip you with the ability to solve this specific integral, but also enhance your overall proficiency in calculus. We will delve into the underlying principles and demonstrate how to apply them effectively. The process involves algebraic manipulation and the application of basic integration rules. By understanding these fundamentals, you can approach a wide range of integration problems with confidence.

Understanding the Integral ∫(2-x)/(1-x) dx

At first glance, the integral ∫(2-x)/(1-x) dx might seem daunting, but it's a classic example of a rational function integration problem. Rational functions, defined as the ratio of two polynomials, often require some algebraic manipulation before they can be integrated directly. In this case, the numerator (2-x) and the denominator (1-x) are both linear polynomials. The key to solving this integral lies in simplifying the integrand, (2-x)/(1-x), into a form that's easier to integrate. This often involves techniques like polynomial long division or algebraic manipulation to separate the function into simpler terms. Understanding the structure of the integrand is crucial for choosing the most appropriate integration strategy. Recognizing the rational form of the function immediately suggests potential approaches, such as long division or partial fraction decomposition. In this specific case, we'll demonstrate how a simple algebraic manipulation can effectively simplify the integral. The goal is to rewrite the integrand in a way that allows us to apply basic integration rules. This often involves separating the function into terms that we can integrate directly, such as constants and simple power functions. By carefully examining the integrand and applying appropriate algebraic techniques, we can transform the integral into a manageable form.

Step-by-Step Solution: Integrating ∫(2-x)/(1-x) dx

To solve the integral ∫(2-x)/(1-x) dx, we'll employ a simple yet effective algebraic manipulation. The goal is to rewrite the numerator in terms of the denominator, allowing us to simplify the fraction. This technique is particularly useful when the degree of the numerator is equal to or greater than the degree of the denominator. Let's begin by adding and subtracting 1 in the numerator: 2 - x = (1 - x) + 1. Now, we can rewrite the integrand as follows: (2 - x) / (1 - x) = ((1 - x) + 1) / (1 - x). Next, we separate the fraction into two terms: ((1 - x) + 1) / (1 - x) = (1 - x) / (1 - x) + 1 / (1 - x). This simplifies to: 1 + 1 / (1 - x). Now, our integral becomes: ∫(2-x)/(1-x) dx = ∫(1 + 1/(1-x)) dx. This form is much easier to integrate. We can now split the integral into two separate integrals: ∫(1 + 1/(1-x)) dx = ∫1 dx + ∫1/(1-x) dx. The first integral, ∫1 dx, is simply x + C₁. For the second integral, ∫1/(1-x) dx, we can use a simple u-substitution. Let u = 1 - x, then du = -dx, and dx = -du. Substituting these into the integral, we get: ∫1/(1-x) dx = ∫1/u (-du) = -∫1/u du. The integral of 1/u is ln|u|, so we have: -∫1/u du = -ln|u| + C₂. Substituting back u = 1 - x, we get: -ln|1 - x| + C₂. Combining the results of both integrals, we have: ∫(2-x)/(1-x) dx = x - ln|1 - x| + C, where C = C₁ + C₂ is the constant of integration. Therefore, the solution to the integral ∫(2-x)/(1-x) dx is x - ln|1 - x| + C. This step-by-step approach demonstrates the power of algebraic manipulation in simplifying integrals. By rewriting the integrand in a more manageable form, we were able to apply basic integration rules and arrive at the solution.

Detailed Integration of ∫1/(1-x) dx Using u-Substitution

As mentioned in the previous step, the integral ∫1/(1-x) dx requires a u-substitution to solve effectively. U-substitution is a powerful technique for simplifying integrals where the integrand contains a composite function. In this case, the composite function is 1/(1-x), where the inner function is (1-x). To apply u-substitution, we first identify a suitable substitution. Let u = 1 - x. This choice is strategic because the derivative of u is a constant multiple of the remaining part of the integrand. Next, we find the derivative of u with respect to x: du/dx = -1. This implies that du = -dx. We can also express dx in terms of du: dx = -du. Now, we substitute u and dx into the integral: ∫1/(1-x) dx = ∫1/u (-du). We can pull the negative sign out of the integral: -∫1/u du. The integral of 1/u with respect to u is the natural logarithm of the absolute value of u: ∫1/u du = ln|u| + C. Therefore, -∫1/u du = -ln|u| + C₂. Finally, we substitute back u = 1 - x to express the result in terms of the original variable x: -ln|u| + C₂ = -ln|1 - x| + C₂. This completes the integration of ∫1/(1-x) dx. This detailed explanation of the u-substitution method highlights its importance in simplifying integrals. By carefully choosing the substitution and following the steps, we can effectively solve a wide range of integrals involving composite functions. The u-substitution technique is a cornerstone of integral calculus and a valuable tool for any student of mathematics.

Alternative Approaches and Considerations

While the algebraic manipulation method we used is efficient, there are alternative approaches to solving ∫(2-x)/(1-x) dx. One such method is polynomial long division. Although not strictly necessary in this case due to the simplicity of the polynomials, it's a useful technique to consider for more complex rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. Polynomial long division would involve dividing (2-x) by (1-x). The result would be a quotient and a remainder, which would then be integrated separately. Another consideration is the domain of the function. The original integrand, (2-x)/(1-x), is undefined when x = 1. This means that the integral is only valid for x ≠ 1. The absolute value in the natural logarithm term, ln|1 - x|, accounts for this restriction, ensuring that the logarithm is always defined. It's crucial to remember the constant of integration, C, when evaluating indefinite integrals. The constant represents the family of functions that have the same derivative as the integrand. Different values of C correspond to vertical shifts of the antiderivative function. In summary, while the algebraic manipulation method is a straightforward approach for this particular integral, exploring alternative methods like polynomial long division and considering the domain and constant of integration provides a more comprehensive understanding of the problem and its solution.

Conclusion: Mastering Rational Function Integration

In conclusion, we've successfully solved the integral ∫(2-x)/(1-x) dx using algebraic manipulation and the technique of u-substitution. We've demonstrated a step-by-step approach that simplifies the integrand into a form that's readily integrable. This process involves rewriting the numerator in terms of the denominator, separating the fraction into simpler terms, and applying the appropriate integration rules. We also explored the u-substitution method in detail, highlighting its importance in integrating composite functions. Furthermore, we discussed alternative approaches, such as polynomial long division, and considerations regarding the domain of the function and the constant of integration. Mastering the integration of rational functions is crucial for a strong foundation in calculus. The techniques discussed in this article, such as algebraic manipulation, u-substitution, and polynomial long division, are applicable to a wide range of integration problems. By understanding these techniques and practicing regularly, you can develop the skills necessary to tackle even the most challenging integrals. Remember to always consider the domain of the function and the constant of integration when evaluating indefinite integrals. With a solid understanding of these concepts, you'll be well-equipped to excel in calculus and related fields.