Question

Multiple choice \(\int_{2}^{3} \frac{3}{(x-1)(x+2)} d x\mathrm{}\) (A) \(-\frac{33}{20}\) (B) \(-\frac{9}{20}\) (C) \(\ln \left(\frac{5}{2}\right)\) (D) \(\ln \left(\frac{8}{5}\right)\) (E) \(\ln \left(\frac{2}{5}\right)\)

Short answer

(D) \(\ln \left(\frac{8}{5}\right)\)

Step-by-step solution

Partial Fraction Decomposition

In order to integrate, we need to break apart the complex fraction into more manageable parts. This is done by expressing the fraction as a sum of simpler fractions or partial fractions. For \(\frac{3}{(x-1)(x+2)}\), it's decomposed as follows: \[ \frac{3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \] By solving this equation, we find the values A and B, which are 1 and -1, respectively.

Substitute and Integrate

Now we substitute our decomposition into the integral: \[ \int_{2}^{3} \frac{3}{(x-1)(x+2)} dx = \int_{2}^{3} (\frac{1}{x-1} - \frac{1}{x+2}) dx \] This breaks our integral into two simpler integrals that we can solve: \[ \int_{2}^{3} \frac{1}{x-1} dx - \int_{2}^{3} \frac{1}{x+2} dx \] We now integrate these fractions, resulting in the natural logarithm. Our integrals become: \[ \ln |x-1| - \ln |x+2| \]

Apply the Limits of Integration and Simplify

With our indefinite integrals calculated, we can now apply the limits of integration to each. This is done by inserting the upper limit and then the lower limit, and subsequently subtracting the two values for each integral. The difference between the two logarithm functions simplifies further to a single logarithm function due to the logarithm properties: \[ \ln \frac{x-1}{x+2} \] Applying the limits gives us: \[ \ln \frac{3-1}{3+2} - \ln \frac{2-1}{2+2} = \ln\frac{2}{5} - \ln\frac{1}{4} = \ln\left(\frac{2}{5}\right) - \ln\left(\frac{1}{4}\right) = \ln\left(\frac{2}{5} \times\frac{4}{1}\right) = \ln\left(\frac{8}{5}\right) \]

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics involved in finding the quantities where the rate of change and the accumulation of quantities is known. It is used to find areas under curves, volumes, central points, and many practical things. In our case, integration is used to calculate the area under the curve of the function \(\frac{3}{(x-1)(x+2)}\) from \(x=2\) to \(x=3\).

Understanding the concept of integrals includes familiarizing oneself with the notation and the process of integration. The integral sign \(\int\) denotes integration, while the dx at the end of the integrand specifies the variable with respect to which we are integrating. The limits of integration, the numbers at the bottom and top of the integral sign, show the range over which we are calculating the area.

Natural Logarithm Properties

The natural logarithm, denoted as \(\ln\), is a mathematical function that is the inverse of the exponential function with base \(e\). The natural logarithm of a number is the power to which \(e\) must be raised to obtain that number. The properties of the natural logarithm play a vital role in simplification during integration.

For instance, \(\ln(ab) = \ln(a) + \ln(b)\) and \(\ln(a/b) = \ln(a) - \ln(b)\) are properties that allow us to simplify the expression after applying the limits of integration. As shown in the solution, by using these properties, we can convert subtraction of two logarithms into a single logarithm of the division of the two original arguments, making it easier to calculate.

Integration Techniques

There are various integration techniques used to solve integrals that are not straightforward. One common technique, particularly for rational functions, is partial fraction decomposition, where you express a complex fraction as a sum of simpler fractions. This method is highly efficient when dealing with quotients of polynomials.

In the provided exercise, we first decomposed the integrand using partial fraction decomposition. This step simplified our complex rational function into two simpler fractions that are easily integratable. It becomes essential to master these techniques as they often are the key to solving integrals that appear unsolvable at first glance.

Limits of Integration

The limits of integration define the interval on the x-axis over which we are integrating. They are incorporated into definite integrals — those that have actual numerical limits, as opposed to indefinite integrals, which do not have specified bounds and include a constant of integration.

When applying limits of integration to indefinite integrals that result from antiderivative functions, like the natural logarithm in our exercise, it involves substituting the upper limit into the antiderivative, subtracting the antiderivative with the lower limit substituted in. This process gave us the final value for our definite integral, leading to the answer \(\ln\left(\frac{8}{5}\right)\). It is crucial to understand that limits of integration help transition from the antiderivative function to the actual area under the curve or the actual quantities we seek to calculate.