Question

In Exercises \(5-14,\) evaluate the integral. $$\int \frac{1-3 x}{3 x^{2}-5 x+2} d x$$

Short answer

The integral is \(ln|3x - 2| - 2 ln|x - 1| + C\)

Step-by-step solution

Partial Fraction Decomposition

The first step is to factor the denominator \(3x^2−5x+2\). This factors into \((3x-2)(x-1)\), so the expression can be rewritten as \(\int \frac{1-3x}{(3x-2)(x-1)} dx\). We want to rewrite this in the form \(\int (\frac{A}{3x-2} + \frac{B}{x-1}) dx\) where A and B are constants to be determined. Multiplying through the denominator, we get the equation \(1-3x = A(x-1) + B(3x-2)\). Now we choose suitable values for x to solve for A and B.

Solve for A and B

If we let \(x = 1\), the equation simplifies to \(1-3 = A(1-1) + B(3-2)\), which simplifies to \(B = -2\). If we let \(x = \frac{2}{3}\), the equation simplifies to \(1 - 2 = A(\frac{2}{3} - 1) + B(3*\frac{2}{3} - 2)\), which simplifies to \(A = 3\). Now we can replace A and B in our equation to get \(\int (\frac{3}{3x-2} - \frac{2}{x-1}) dx\)

Integration

Now we can integrate each term separately: \(\int (\frac{3}{3x-2} - \frac{2}{x-1}) dx = 3 \int \frac{1}{3x-2} dx - 2 \int \frac{1}{x-1} dx\). The integrals are standard forms, so we easily find: \(ln|3x - 2| - 2 ln|x - 1| + C\).

Simplify and Write Down Final Answer

In the end, the integral is \(ln|3x - 2| - 2 ln|x - 1| + C\)

Key concepts

The Indefinite Integral

The concept of the indefinite integral, broadly referred to as antiderivative, is central to calculus. It represents the general form of all the antiderivatives of a function. Essentially, when we are finding the indefinite integral of a function, we're determining a function whose derivative is the given function. In notation, the indefinite integral of a function f(x) with respect to x is written as \(\int f(x) \, dx\) and includes a constant of integration, often denoted by C. This constant appears because the process of differentiation ignores any constant term (as its derivative is zero).

In our example, evaluating \(\int \frac{1-3 x}{3 x^{2}-5 x+2} \, dx\) is about reversing the differentiation process. The result is an expression involving a natural logarithm, as logarithmic functions are the antiderivatives of reciprocal functions, and a plus C to account for the constant of integration that could be part of the original function before it was derived.

Integration Techniques

Mastering various integration techniques is essential for solving complex integrals that cannot be evaluated by basic methods. Some standard techniques include substitution, integration by parts, and partial fraction decomposition. Our focus here is on partial fraction decomposition, a method used when integrating rational functions (ratios of polynomials). It involves expressing a complex rational function as a sum of simpler fractions whose denominators are factors of the original denominator.

For the given function \(\frac{1-3 x}{3 x^{2}-5 x+2}\), we first factor the denominator and then represent the integral as a sum of fractions with unknown coefficients (in our case A and B). By solving for A and B, the integral is broken down into simpler parts that are easier to evaluate independently. Factoring, a key algebraic manipulation, simplifies the integration process and lets us deal with each fraction separately, leading to an outcome involving simpler functions like logarithms.

Algebraic Manipulation for Integration

Algebraic manipulation plays a pivotal role in the preparation for integration, particularly when using partial fraction decomposition. To successfully integrate by this method, we must first manipulate the given algebraic expression to a suitable form, as shown in our example. After decomposing into partial fractions: \(\frac{1-3x}{(3x-2)(x-1)} dx = \int (\frac{A}{3x-2} + \frac{B}{x-1}) dx\), we equate this to the original numerator and solve for the constants A and B through strategic choices of x values.

This step requires skillful algebra, such as expanding, simplifying, and comparing coefficients. These algebraic steps are crucial as they transform the integral into a more manageable form. Once A and B are found, in our function being A = 3 and B = -2, simpler integrals emerge, which can be straightforwardly evaluated using basic antiderivatives. This highlights the importance of algebra in making complex integrals accessible.