Evaluating Integral Of (7x - 13) / ((3x - 5)(x - 3)) Using Partial Fractions

Introduction

In this comprehensive article, we will delve into the step-by-step evaluation of the integral ∫ (7x - 13) / ((3x - 5)(x - 3)) dx. This type of integral, involving a rational function, is a common challenge in calculus, often requiring the technique of partial fraction decomposition. Our goal is to provide a clear, detailed explanation of each step, ensuring that readers can confidently tackle similar problems. We will begin by outlining the foundational principles of partial fraction decomposition, then apply these principles to our specific problem, and finally, integrate the resulting terms to arrive at the solution.

Understanding Partial Fraction Decomposition

The core technique we'll employ is partial fraction decomposition. This method is indispensable for integrating rational functions where the denominator can be factored. A rational function is simply a ratio of two polynomials, and when its denominator is a product of simpler polynomials, we can decompose the original fraction into a sum of simpler fractions. Each of these simpler fractions has one of the factors from the original denominator as its denominator, and a constant (or a simpler polynomial) as its numerator. This decomposition makes the integration process significantly easier because the simpler fractions are usually straightforward to integrate.

The fundamental principle behind partial fraction decomposition is to rewrite a complex rational expression as a sum of simpler fractions. This technique is particularly useful when the denominator of the rational expression can be factored into linear or quadratic factors. For example, if we have a rational function of the form P(x) / Q(x), where Q(x) can be factored into (ax + b)(cx + d), we can express the original fraction as A / (ax + b) + B / (cx + d), where A and B are constants. The challenge then becomes finding the values of these constants.

To determine the constants, we combine the decomposed fractions back into a single fraction and equate the numerators. This results in an algebraic equation that we can solve for the unknown constants. Once we have these constants, the original integral is transformed into a sum of simpler integrals, each of which is typically easier to evaluate. In our case, the denominator (3x - 5)(x - 3) is already factored into linear terms, making partial fraction decomposition an ideal approach.

Setting up the Partial Fraction Decomposition

To begin, we recognize that the integrand (7x - 13) / ((3x - 5)(x - 3)) is a rational function suitable for partial fraction decomposition. The denominator is already factored into two distinct linear factors: (3x - 5) and (x - 3). Therefore, we can express the integrand as a sum of two fractions, each with one of these factors as its denominator. We introduce constants A and B as the numerators of these fractions, giving us the following equation:

(7x - 13) / ((3x - 5)(x - 3)) = A / (3x - 5) + B / (x - 3)

The next step is to find the values of the constants A and B. To do this, we clear the denominators by multiplying both sides of the equation by the original denominator, (3x - 5)(x - 3). This gives us:

7x - 13 = A(x - 3) + B(3x - 5)

This equation is now free of fractions, making it easier to manipulate and solve for A and B. We can expand the right side of the equation to get:

7x - 13 = Ax - 3A + 3Bx - 5B

Now, we group the terms with x and the constant terms:

7x - 13 = (A + 3B)x + (-3A - 5B)

For this equation to hold true for all values of x, the coefficients of x on both sides must be equal, and the constant terms on both sides must be equal. This gives us a system of two linear equations:

A + 3B = 7 -3A - 5B = -13

We now have a system of two equations with two unknowns, which we can solve using various methods, such as substitution or elimination. Solving this system will give us the values of A and B, which are crucial for completing the partial fraction decomposition.

Solving for the Constants A and B

Continuing from the system of equations we derived earlier:

A + 3B = 7 -3A - 5B = -13

We can use the method of elimination to solve for A and B. First, we multiply the first equation by 3 to make the coefficients of A in both equations additive inverses:

3(A + 3B) = 3(7) 3A + 9B = 21

Now we have the following system:

3A + 9B = 21 -3A - 5B = -13

Adding the two equations eliminates A:

(3A + 9B) + (-3A - 5B) = 21 + (-13) 4B = 8

Dividing both sides by 4 gives us the value of B:

B = 8 / 4 B = 2

Now that we have the value of B, we can substitute it back into one of the original equations to solve for A. We'll use the first equation, A + 3B = 7:

A + 3(2) = 7 A + 6 = 7

Subtracting 6 from both sides gives us the value of A:

A = 7 - 6 A = 1

So, we have found that A = 1 and B = 2. These values are essential for rewriting the original integrand using partial fraction decomposition.

Rewriting the Integral

With the values of A and B determined (A = 1 and B = 2), we can now rewrite the original integrand using the partial fraction decomposition. Recall that we expressed the integrand as:

(7x - 13) / ((3x - 5)(x - 3)) = A / (3x - 5) + B / (x - 3)

Substituting the values of A and B, we get:

(7x - 13) / ((3x - 5)(x - 3)) = 1 / (3x - 5) + 2 / (x - 3)

This decomposition transforms the original integral into a sum of two simpler integrals:

∫ (7x - 13) / ((3x - 5)(x - 3)) dx = ∫ [1 / (3x - 5) + 2 / (x - 3)] dx

We can now split the integral into two separate integrals:

∫ [1 / (3x - 5) + 2 / (x - 3)] dx = ∫ 1 / (3x - 5) dx + ∫ 2 / (x - 3) dx

These two integrals are much easier to evaluate than the original integral. The next step is to integrate each of these terms separately.

Integrating the Decomposed Fractions

Now, let's integrate each term separately. We have two integrals to evaluate:

∫ 1 / (3x - 5) dx and ∫ 2 / (x - 3) dx

For the first integral, ∫ 1 / (3x - 5) dx, we can use a simple u-substitution. Let u = 3x - 5. Then, du = 3 dx, so dx = (1/3) du. Substituting these into the integral, we get:

∫ 1 / (3x - 5) dx = ∫ (1/u) (1/3) du = (1/3) ∫ (1/u) du

The integral of 1/u with respect to u is the natural logarithm of the absolute value of u:

(1/3) ∫ (1/u) du = (1/3) ln|u| + C1

Substituting back for u, we get:

(1/3) ln|3x - 5| + C1

Now, let's evaluate the second integral, ∫ 2 / (x - 3) dx. This integral is also straightforward. We can rewrite it as:

∫ 2 / (x - 3) dx = 2 ∫ 1 / (x - 3) dx

The integral of 1 / (x - 3) with respect to x is the natural logarithm of the absolute value of (x - 3):

2 ∫ 1 / (x - 3) dx = 2 ln|x - 3| + C2

So, we have evaluated both integrals. The next step is to combine these results to find the final answer.

Combining the Results

We have found the integrals of the decomposed fractions:

∫ 1 / (3x - 5) dx = (1/3) ln|3x - 5| + C1 ∫ 2 / (x - 3) dx = 2 ln|x - 3| + C2

Now, we add these results together to find the integral of the original expression:

∫ (7x - 13) / ((3x - 5)(x - 3)) dx = (1/3) ln|3x - 5| + 2 ln|x - 3| + C

Here, C represents the constant of integration, which combines C1 and C2. This is the final result of the integration. We have successfully evaluated the integral using partial fraction decomposition and basic integration techniques.

Conclusion

In this article, we have thoroughly evaluated the integral ∫ (7x - 13) / ((3x - 5)(x - 3)) dx. We began by introducing the method of partial fraction decomposition, a crucial technique for integrating rational functions. We then applied this method to the given integrand, breaking it down into simpler fractions. We solved for the constants in the decomposition, rewrote the integral as a sum of simpler integrals, and evaluated each of these integrals. Finally, we combined the results to obtain the solution:

∫ (7x - 13) / ((3x - 5)(x - 3)) dx = (1/3) ln|3x - 5| + 2 ln|x - 3| + C

This detailed walkthrough illustrates the power and utility of partial fraction decomposition in calculus. By understanding and applying this technique, you can tackle a wide range of integrals involving rational functions. Remember to always factor the denominator, set up the partial fractions correctly, solve for the constants, and then integrate each term. With practice, this method will become a valuable tool in your calculus toolkit.