Solving ∫ (5x^2 - 14x + 3) / ((x-1)(x-3)^2) Dx
Hey guys! Let's dive into a fun math problem today. We're going to tackle the indefinite integral of a rational function. Specifically, we're looking at how to solve:
∫ (5x^2 - 14x + 3) / ((x-1)(x-3)^2) dx
This might look intimidating at first, but don't worry! We'll break it down step by step using a technique called partial fraction decomposition. This method is super useful for integrating rational functions, which are just fractions where the numerator and denominator are polynomials. So, grab your pencils, and let's get started!
Understanding Partial Fraction Decomposition
Before we jump into the actual integral, let's quickly recap what partial fraction decomposition is all about. The main idea is to break down a complex rational fraction into simpler fractions that are easier to integrate. Think of it like taking a complicated recipe and breaking it down into individual ingredients – much easier to handle that way, right?
Partial fraction decomposition is a powerful technique used to simplify rational functions, making them easier to integrate. The core idea is to express a complex fraction as a sum of simpler fractions. For our integral, this means rewriting the expression (5x^2 - 14x + 3) / ((x-1)(x-3)^2) as a sum of fractions with denominators (x-1) and (x-3). This method hinges on the principle that any rational function where the degree of the numerator is less than the degree of the denominator can be decomposed into a sum of fractions with simpler denominators. The key is to identify the correct form of the decomposition based on the factors in the original denominator. In our case, we have a linear factor (x-1) and a repeated linear factor (x-3)^2, which will guide us in setting up the partial fractions. By breaking down the complex fraction into simpler parts, we transform a challenging integral into a series of more manageable integrals, significantly simplifying the problem-solving process. This technique is not only applicable to integration but also useful in various areas of mathematics and engineering where rational functions are encountered.
Setting Up the Partial Fractions
Alright, so the first thing we need to do is set up the partial fraction decomposition. Looking at our integral, we see the denominator has two factors: (x-1) and (x-3)^2. Because (x-3) is squared, we need to include both (x-3) and (x-3)^2 in our decomposition. This gives us the following form:
(5x^2 - 14x + 3) / ((x-1)(x-3)^2) = A / (x-1) + B / (x-3) + C / (x-3)^2
Here, A, B, and C are constants that we need to find. Think of them as the secret ingredients we need to unlock the solution! The goal here is to find the values of A, B, and C that make this equation true for all values of x (except for x=1 and x=3, where the original fraction is undefined).
Setting up partial fractions correctly is crucial for the success of this integration method. The form of the decomposition is dictated by the factors in the denominator of the original rational function. In our specific case, the denominator (x-1)(x-3)^2 contains a linear factor (x-1) and a repeated linear factor (x-3)^2. This necessitates including three partial fractions in our decomposition: A / (x-1), B / (x-3), and C / (x-3)^2. Each factor in the denominator corresponds to a partial fraction, and the repeated factor (x-3)^2 requires two terms to account for its multiplicity. The constants A, B, and C are placeholders for the unknown coefficients that we need to determine. These constants will allow us to express the original complex fraction as a sum of simpler fractions, each of which is easier to integrate. By carefully setting up the partial fractions according to the factors in the denominator, we lay the foundation for solving the integral using straightforward integration techniques. This initial setup is a critical step in the process, as it transforms the problem into a manageable form suitable for further algebraic manipulation and integration.
Solving for A, B, and C
Now comes the fun part – solving for those constants! To do this, we'll multiply both sides of the equation by the original denominator, (x-1)(x-3)^2. This clears out the fractions and leaves us with a polynomial equation:
5x^2 - 14x + 3 = A(x-3)^2 + B(x-1)(x-3) + C(x-1)
We can solve for A, B, and C using a couple of different methods. One way is to expand the right side, collect like terms, and then equate the coefficients of the corresponding powers of x on both sides. This gives us a system of linear equations that we can solve. Another, often quicker, way is to strategically choose values of x that will make some of the terms zero, allowing us to solve for the constants one at a time.
Let's use the second method. First, let's set x = 1. This makes the terms with (x-1) equal to zero, leaving us with:
5(1)^2 - 14(1) + 3 = A(1-3)^2
5 - 14 + 3 = 4A
-6 = 4A
A = -3/2
Great! We found A. Now, let's set x = 3. This makes the terms with (x-3) equal to zero, giving us:
5(3)^2 - 14(3) + 3 = C(3-1)
45 - 42 + 3 = 2C
6 = 2C
C = 3
Awesome, we've got C too! To find B, we can choose any other value of x. A simple one is x = 0. Plugging that in, along with the values we found for A and C, we get:
5(0)^2 - 14(0) + 3 = A(0-3)^2 + B(0-1)(0-3) + C(0-1)
3 = 9A + 3B - C
3 = 9(-3/2) + 3B - 3
3 = -27/2 + 3B - 3
9 = -27/2 + 3B
3B = 9 + 27/2
3B = 45/2
B = 15/2
So, we've found A = -3/2, B = 15/2, and C = 3. We've cracked the code!
Solving for the constants A, B, and C is a crucial step in the partial fraction decomposition process. We have several methods at our disposal to accomplish this, each with its own advantages. One common approach is to multiply both sides of the equation by the original denominator, which clears the fractions and leaves us with a polynomial equation. From there, we can either expand the right side, collect like terms, and equate coefficients of corresponding powers of x, or we can strategically choose values of x that simplify the equation. The method of equating coefficients involves setting up a system of linear equations based on the coefficients of the polynomial terms and then solving that system. Alternatively, the method of strategic substitution involves choosing values of x that make certain terms zero, allowing us to isolate and solve for the constants one at a time. In our case, we employed the strategic substitution method, which proved to be efficient. By substituting x = 1 and x = 3, we were able to directly solve for A and C, respectively. To find B, we substituted x = 0 along with the known values of A and C. This approach significantly simplifies the process and allows us to quickly determine the values of the constants, paving the way for the integration step.
Rewriting the Integral
Now that we have our constants, we can rewrite the original integral using the partial fraction decomposition:
∫ (5x^2 - 14x + 3) / ((x-1)(x-3)^2) dx = ∫ (-3/2) / (x-1) + (15/2) / (x-3) + 3 / (x-3)^2 dx
This looks much more manageable, doesn't it? We've transformed a complex integral into a sum of simpler integrals. We can now split this up into three separate integrals:
= (-3/2) ∫ 1 / (x-1) dx + (15/2) ∫ 1 / (x-3) dx + 3 ∫ 1 / (x-3)^2 dx
Rewriting the integral using the determined constants is the pivotal step that transforms the original complex integral into a form that is readily integrable. By substituting the values of A, B, and C into the partial fraction decomposition, we express the integrand as a sum of simpler fractions. This decomposition allows us to split the original integral into a sum of integrals, each of which corresponds to one of the partial fractions. In our case, we replaced (5x^2 - 14x + 3) / ((x-1)(x-3)^2) with (-3/2) / (x-1) + (15/2) / (x-3) + 3 / (x-3)^2. This transformation significantly simplifies the integration process because each of these resulting fractions is easier to integrate than the original complex fraction. We now have three separate integrals to evaluate: (-3/2) ∫ 1 / (x-1) dx, (15/2) ∫ 1 / (x-3) dx, and 3 ∫ 1 / (x-3)^2 dx. Each of these integrals can be solved using basic integration techniques, such as recognizing the antiderivative of 1/u as ln|u| and applying the power rule to the integral of 1/(x-3)^2. The key benefit of this rewriting step is that it breaks down a formidable problem into a series of manageable tasks, making the overall integration process straightforward and accessible.
Evaluating the Integrals
Now, let's evaluate each of these integrals. The first two are straightforward logarithmic integrals, and the third one can be solved using a simple power rule substitution.
∫ 1 / (x-1) dx = ln|x-1| + C_1
∫ 1 / (x-3) dx = ln|x-3| + C_2
For the third integral, let's use a substitution. Let u = x - 3, so du = dx. Then the integral becomes:
∫ 1 / (x-3)^2 dx = ∫ 1 / u^2 du = ∫ u^(-2) du = -u^(-1) + C_3 = -1 / (x-3) + C_3
Evaluating the individual integrals is the culmination of the partial fraction decomposition method, where we finally find the antiderivatives of the simplified fractions. The first two integrals, (-3/2) ∫ 1 / (x-1) dx and (15/2) ∫ 1 / (x-3) dx, are classic examples of integrals that yield natural logarithms. Recognizing that the antiderivative of 1/u with respect to u is ln|u| + C, we quickly find the integrals to be (-3/2)ln|x-1| and (15/2)ln|x-3|, respectively. The absolute value ensures that the logarithm is defined for all x except x = 1 and x = 3. For the third integral, 3 ∫ 1 / (x-3)^2 dx, we employ a simple u-substitution to transform it into a more manageable form. By letting u = x - 3, we have du = dx, and the integral becomes ∫ 1 / u^2 du. Rewriting 1 / u^2 as u^(-2), we apply the power rule for integration, which states that ∫ u^n du = (u^(n+1)) / (n+1) + C, for n ≠ -1. Thus, we find the integral to be -u^(-1) + C, which we then rewrite as -1 / (x-3) after substituting back for u. Each of these integrals is straightforward to evaluate, highlighting the power of partial fraction decomposition in transforming a complex integral into a sum of simpler ones. This step-by-step evaluation ensures we accurately determine the antiderivatives, setting the stage for the final combination of terms to arrive at the solution.
Combining the Results
Finally, let's plug these results back into our equation and combine them. Remember to include the constant of integration, C, which accounts for the arbitrary constant that can be added to any antiderivative.
∫ (5x^2 - 14x + 3) / ((x-1)(x-3)^2) dx = (-3/2)ln|x-1| + (15/2)ln|x-3| - 3 / (x-3) + C
And there we have it! We've successfully evaluated the indefinite integral.
Combining the results is the final step in our journey, where we piece together the antiderivatives of the individual partial fractions to form the solution to the original integral. We substitute the evaluated integrals back into our equation, remembering to include the constants of integration for each term. These constants are then consolidated into a single constant of integration, C, which represents the family of antiderivatives for the given function. In our case, we have:
∫ (5x^2 - 14x + 3) / ((x-1)(x-3)^2) dx = (-3/2)ln|x-1| + (15/2)ln|x-3| - 3 / (x-3) + C
This final expression represents the indefinite integral of the original rational function. It combines the logarithmic terms arising from the linear factors and the rational term resulting from the repeated linear factor. The constant C accounts for the fact that the derivative of a constant is zero, so there are infinitely many antiderivatives that differ only by a constant. This step not only provides the complete solution but also underscores the elegance and effectiveness of the partial fraction decomposition method. By carefully breaking down the complex rational function into simpler components and then integrating each component separately, we arrive at a solution that would have been challenging to obtain directly. The final combination of results is a testament to the power of this technique in simplifying integration problems.
Conclusion
So, guys, we've successfully navigated through a complex integral using partial fraction decomposition. It might seem like a lot of steps, but each step is logical and manageable. Remember, the key is to break down the problem into smaller, easier-to-solve parts. Keep practicing, and you'll become a master of integration in no time!
