91Ó°ÊÓ

When faced with the challenge of integrating a complex rational function, one of the most powerful techniques we can use is partial fraction decomposition. This method involves breaking down a complicated fraction into a sum of simpler fractions. The goal is to rewrite the integrand in a form that's easier to integrate.

In this case, we start with a rational function with a polynomial numerator (-12x² - x + 33) and a cubic polynomial in the denominator ((x-1)(x+3)(3-2x)). Using partial fractions, we express this complex fraction as a sum of three simpler fractions:

• \( \frac{A}{x-1} \), where \( A \) is a constant associated with the factor \( x-1 \).
• \( \frac{B}{x+3} \), where \( B \) corresponds to the factor \( x+3 \).
• \( \frac{C}{3-2x} \), where \( C \) is linked with \( 3-2x \).

To determine these constants \( A \), \( B \), and \( C \), we clear the denominators by multiplying both sides by ((x-1)(x+3)(3-2x)). We then expand the right-side expression and collect like terms. By equating coefficients from both sides, we can form a system of linear equations. Solving this system gives us the values of the constants. It's a methodical process that paves the way for easier integration.

Once we've broken down a rational function into partial fractions, we can move on to actually integrating these simpler components. Rational function integration involves straightforward calculus techniques once you have the function in an easier-to-handle form.

The integral \(\int \frac{-12x^2 - x + 33}{(x-1)(x+3)(3-2x)} dx\) transforms into \(\int \left( \frac{2}{x-1} - \frac{3}{x+3} - \frac{1}{3-2x} \right) dx\) using the previously found constants \( A = 2 \), \( B = -3 \), and \( C = -1 \). Now, the integration becomes much simpler:

• Integrate \( \frac{2}{x-1} \) to get \( 2 \ln|x-1| \).
• Integrate \( -\frac{3}{x+3} \) to get \(-3 \ln|x+3| \).
• Integrate \( -\frac{1}{3-2x} \), and apply a u-substitution if needed, to get \( \frac{1}{2} \ln|3-2x| \).

Each part is integrated using basic rules of logarithmic integration. This step is key to turning a seemingly intractable problem into a manageable series of calculations.

Logarithmic integration is a valuable technique when dealing with integrals of the form \( \frac{k}{x+a} \), where \( k \) is a constant and \( x+a \) is a linear function of \( x \). The result of integrating such a term is a natural logarithm. This is a simple yet powerful method in calculus.

In our exercise, after applying partial fraction decomposition, we integrate terms like \( \frac{2}{x-1} \), \( \frac{-3}{x+3} \), and \( \frac{-1}{3-2x} \). These are perfect candidates for logarithmic integration, which results in natural logarithm functions:

• \( \int \frac{2}{x-1} \, dx \rightarrow 2 \ln|x-1| \)
• \( \int \frac{-3}{x+3} \, dx \rightarrow -3 \ln|x+3| \)
• \( \int \frac{-1}{3-2x} \, dx \rightarrow \frac{1}{2} \ln|3-2x| \)

The addition of these logarithmic results gives the final integral. Importantly, don't forget to include a constant of integration \( C \) at the end of indefinite integrals, acknowledging that there are infinitely many antiderivatives.