Question

Evaluate the following integrals: $$ \int \frac{x}{x^{2}-5 x+6} d x $$

Short answer

Question: Determine the antiderivative of the function \(f(x) = \frac{x}{x^2 - 5x + 6}\). Answer: The antiderivative of the function \(f(x) = \frac{x}{x^2 - 5x + 6}\) is given by \(-2\ln(|x-2|) + 3\ln(|x-3|) + C\), where C is the constant of integration.

Step-by-step solution

Factor the denominator

First, we need to factor the denominator \(x^2 - 5x + 6\). Since the coefficients are fairly small, we can guess and check possible factors or use the quadratic formula to find the roots of the quadratic equation. In this case, we see that the denominator can be factored as \((x-3)(x-2)\). So, the integrand can be written as: $$ \frac{x}{(x-3)(x-2)} $$

Perform partial fraction decomposition

Now, we need to rewrite the integrand as a sum of simpler fractions with denominators \((x-3)\) and \((x-2)\). We set up the relation: $$ \frac{x}{(x-3)(x-2)} = \frac{A}{x-2} + \frac{B}{x-3} $$ Multiplying both sides by \((x-3)(x-2)\) to clear the fractions, we get: $$ x = A(x-3) + B(x-2) $$

Find A and B

Now, we need to find the constants A and B. To do that, we can use the cover-up method or simply substitute values of x and solve for A and B. Using the cover-up method, we cover up \((x-2)\) in the denominator of \(\frac{x}{(x-3)(x-2)}\) and substitute x = 2 into the remaining expression: $$ A = \frac{2}{(2-3)} = -2 $$ Similarly, cover up \((x-3)\) in the denominator of \(\frac{x}{(x-3)(x-2)}\) and substitute x = 3 into the remaining expression: $$ B = \frac{3}{(3-2)} = 3 $$ Now, we can rewrite the integrand with our found values of A and B: $$ \frac{x}{(x-3)(x-2)} = \frac{-2}{x-2} + \frac{3}{x-3} $$

Integrate term by term

Finally, we can now integrate each term with respect to x: $$ \int \frac{x}{x^2 - 5x + 6} dx = \int \left(\frac{-2}{x-2} + \frac{3}{x-3}\right) dx $$ Now, integrate each term: $$ = -2\int \frac{1}{x-2} dx + 3\int \frac{1}{x-3} dx $$ The integral of \(\frac{1}{x-a}\) is \(\ln(|x-a|)\), so we get: $$ =-2\ln(|x-2|) + 3\ln(|x-3|) + C $$ Where C is the constant of integration. So, the final solution is: $$ \int \frac{x}{x^{2}-5 x+6} d x = -2\ln(|x-2|) + 3\ln(|x-3|) + C $$

Key concepts

Partial Fraction Decomposition

Understanding partial fraction decomposition is crucial for solving complex integral problems involving rational functions. Picture a complex fraction, like a jigsaw puzzle, that needs to be broken down into simpler, more manageable pieces. This is what partial fraction decomposition achieves.

For the integral at hand, we apply this technique to the integrand \(\frac{x}{x^{2}-5x+6}\). By factoring the denominator—splitting \(x^{2}-5x+6\) into \(x-3\) and \(x-2\) after recognizing it as a product of two linear factors—we can then represent the original fraction as a sum of fractions with these linear factors in their denominators. Assign letters like \(A\) and \(B\) to the numerators of these new fractions: \(\frac{A}{x-2}\) and \(\frac{B}{x-3}\), respectively.

Afterwards, you'll find the values of \(A\) and \(B\) through smart substitutions or comparing coefficients, effectively simplifying the integrand into components that are much easier to integrate. This method not only aids in performing the integration but also provides a systematic approach to tackle seemingly daunting integrals.

Factor the Denominator

Factoring the denominator is a step that often precedes partial fraction decomposition. It's like finding the right tools before disassembling a machine. For instance, the denominator \(x^{2}-5x+6\) in our integral may initially look impenetrable. However, with a keen eye, one can recognize that it's a quadratic polynomial that can be broken down into the product of two binomials.

Typically, we look for two numbers that multiply to give the constant term (+6 in our case) and add up to the linear coefficient (-5 in this example). Here, the numbers -3 and -2 fit the bill. Thus, we rewrite the original quadratic as \( (x-3)(x-2)\). Factoring is a skill that simplifies many steps in integral calculus, making it easier to apply partial fraction decomposition, which then facilitates the integration process.

Integration by Logarithms

Once we've decomposed the fraction and factored the denominator, we often encounter integrals of the form \(\int \frac{1}{x-a} dx\), which are tailor-made for integration by logarithms. Think of it like taking the 'signature' of the function: the natural logarithm captures the essence of these types of integrals.

In our exercise, after finding the values of \(A\) and \(B\), we arrive at terms \(\frac{-2}{x-2}\) and \(\frac{3}{x-3}\). Each of these can be directly integrated using logarithms. The antiderivative of \(\frac{1}{x-a}\) is \(\ln(|x-a|)\), resulting in an answer that incorporates natural logarithms of absolute values:

• \( -2\ln(|x-2|) \)
• \( +3\ln(|x-3|) \)

Adding a constant of integration \(C\) completes the process. This method is incredibly powerful in its simplicity and is a testament to how logarithms are seamlessly interwoven with calculus.