Problem 70 - Calculus for Biology and Medicine

Question

Evaluate each integral. $$ \int \frac{2 x-1}{(x+4)(x+1)} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 3\ln |x+4| - \ln |x+1| + C \).

Step 1: Recognize the Method

Notice that the integrand is a rational function where the degree of the numerator is less than the degree of the denominator. This suggests that partial fraction decomposition will be useful.

Step 2: Set Up Partial Fractions

Express the fraction \( \frac{2x-1}{(x+4)(x+1)} \) as the sum of two simpler fractions: \( \frac{A}{x+4} + \frac{B}{x+1} \).

Step 3: Solve for the Constants

Multiply through by the denominator \((x+4)(x+1)\) to clear the fractions and create an equation: \( 2x - 1 = A(x + 1) + B(x + 4) \). Expand and combine like terms to get: \( 2x - 1 = Ax + A + Bx + 4B \).

Step 4: Rearrange and Collect Like Terms

Rearrange to get \( 2x - 1 = (A + B)x + (A + 4B) \). Compare coefficients with the original numerator \(2x - 1\): \( A + B = 2 \) and \( A + 4B = -1 \).

Step 5: Solve the System of Equations

Solve the system of equations: 1. \( A + B = 2 \) 2. \( A + 4B = -1 \)Subtract equation 1 from equation 2 to solve for \(B\), yielding \(3B = -3\), so \(B = -1\). Substitute \(B = -1\) back into equation 1 to find \(A = 3\).

Step 6: Rewrite the Integral with Partial Fractions

Using \(A = 3\) and \(B = -1\), rewrite the integral: \( \int \left( \frac{3}{x+4} - \frac{1}{x+1} \right) dx \).

Step 7: Integrate Each Term

Integrate each fraction separately:1. \( \int \frac{3}{x+4} \, dx = 3 \ln |x+4| + C_1 \)2. \( -\int \frac{1}{x+1} \, dx = -\ln |x+1| + C_2 \) Combine the constants \(C_1\) and \(C_2\) into a single constant \(C\).

Step 8: Write the Final Solution

The integral evaluates to: \[ 3\ln |x+4| - \ln |x+1| + C \].

Key Concepts

Integrals

In calculus, integrals are fundamental concepts that provide a way to calculate areas under curves, among other things. They allow you to accumulate quantities and often represent the total of continuously distributed quantities.
Integrals can be indefinite or definite:

• Indefinite integrals represent a family of functions and include a constant of integration \( C \). It is expressed as \( \int f(x) \, dx \) and results in an antiderivative of \( f(x) \).
• Definite integrals calculate the net area under a curve between two limits, \( a \) and \( b \), and are represented by \( \int_a^b f(x) \, dx \).

The problem in the exercise deals with an indefinite integral requiring partial fraction decomposition before integrating. This simplifies the process, especially for rational functions.

Rational Functions

Rational functions are fractions where the numerator and the denominator are both polynomials. They are expressed in the form \( \frac{P(x)}{Q(x)} \). An essential feature of rational functions is the relationship between the degrees of the polynomials:

• If the degree of the numerator preceeds that of the denominator, polynomial long division can simplify the function first.
• If the degree of the numerator is less than the denominator, partial fraction decomposition aids in simplifying the fractional expression, as seen in the provided exercise.

In our scenario, the integrand \( \frac{2x-1}{(x+4)(x+1)} \) is such that the degree of the numerator is lower than the denominator, indicating the suitability of partial fraction decomposition. This technique rewrites the function as a sum of simpler rational terms.

System of Equations

A system of equations involves multiple equations with common variables. Solving such a system means finding the values of variables that satisfy all equations simultaneously.
In partial fraction decomposition, determining the constants requires setting up a system of equations. From the exercise, the system was:

• \( A + B = 2 \)
• \( A + 4B = -1 \)

This linear system can be solved using methods such as substitution, elimination, or matrix operations. Here, we used elimination to see that solving for \( A \) and \( B \) involved subtracting one equation from the other to isolate a variable. This simplification process eventually determined the values of \( A \) and \( B \), simplifying the integral into separate terms that could be easily evaluated.