Problem 44 - Calculus for Biology and Medicine

Question

Use partial fraction decompositions to evaluate each integral. $$ \int \frac{2 x^{2}-2 x+1}{x^{2}(x-1)} d x $$

Step-by-Step Solution

Answer

\( \int \frac{2x^2 - 2x + 1}{x^2(x-1)} \, dx = \ln|x| - \frac{1}{x} + \ln|x-1| + C \)

Step 1: Set up Partial Fractions

The integrand \( \frac{2x^2 - 2x + 1}{x^2(x-1)} \) can be decomposed into partial fractions. We express it as \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} \).

Step 2: Clear the Denominator

Multiply both sides by \( x^2(x-1) \) to clear the denominator: \[ 2x^2 - 2x + 1 = A x (x-1) + B (x-1) + C x^2 \].

Step 3: Expand and Simplify

Expand the right side: \( A x^2 - A x + B x - B + C x^2 \) which simplifies to \( (A+C)x^2 + (B-A)x - B \).

Step 4: Match Coefficients

Match coefficients with \( 2x^2 - 2x + 1 \), leading to: 1. \( A + C = 2 \) 2. \( B - A = -2 \) 3. \( -B = 1 \).

Step 5: Solve the System of Equations

Solve the system:- From \(-B = 1\), get \( B = -1\).- From \( B - A = -2 \), substitute \( B = -1 \) to get \(-1 - A = -2\) which gives \( A = 1 \).- From \( A + C = 2\), substitute \( A = 1 \) to get \( 1 + C = 2 \) which gives \( C = 1 \).

Step 6: Write the Integral Expression

The fraction deconstruction is confirmed as: \( \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} \). Thus, the integral becomes: \[ \int \left( \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} \right) \, dx \].

Step 7: Integrate Term by Term

Integrate each term separately:\[ \int \frac{1}{x} \, dx = \ln|x| + C_1 \]\[ \int -\frac{1}{x^2} \, dx = \frac{1}{x} + C_2 \]\[ \int \frac{1}{x-1} \, dx = \ln|x-1| + C_3 \].

Step 8: Combine the Integrals

Combine the results to obtain the final solution:\[ \ln|x| - \frac{1}{x} + \ln|x-1| + C \].

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used in calculus for breaking down complex rational expressions into simpler fractions. This technique can be particularly handy when dealing with integrals that involve polynomials in the denominator, as seen in the exercise.

Here's the key idea: take an expression like \( \frac{2x^2 - 2x + 1}{x^2(x-1)} \). Notice that the denominator has already been factored into \( x^2(x-1) \), making it a perfect candidate for partial fractions. We then hypothesize that this complex fraction can be split into a sum of simpler fractions: \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} \).

The process consists of these steps:

• Clear the denominators by multiplying through by the least common denominator, which gives us simple polynomials to work with on both sides.
• Expand the equation, if necessary, to reveal all terms.
• Match coefficients from both sides of the equation in order to find the values for \( A, B, \) and \( C \).
• Solve this system of linear equations to find the unknown constants.

By simplifying complex expressions in this way, integration becomes much easier.

Integral Calculus

Integral calculus is one of the two main branches of calculus, the other being differential calculus. In integral calculus, we are concerned with the summation of quantities over a continuous range. That means calculating the integral of functions.

Integration is essentially the reverse operation of differentiation. It can be applied to compute areas under curves, the total accumulated change, and other quantities of interest in mathematics and applied sciences. Fundamentally, integration finds the scale or total based on a rate of change, while differentiation determines that rate of change from a total.

In our exercise, the challenge is to solve an integral of the form \( \int \frac{2x^2 - 2x + 1}{x^2(x-1)} \, dx \), where using the partial fraction decomposition method makes each part manageable. Integrals like these, when written in broken-down simple fractions, are easier to evaluate using basic integral formulas.

Mathematical Integration

Mathematical integration portrays the process of finding the antiderivative or the area under the curve of a function. The basic indefinite integrals that arise in calculus problems can often be solved using known formulas.

Once your fraction is decomposed, each term, such as \( \frac{1}{x} \), \( \frac{1}{x^2} \), and \( \frac{1}{x-1} \), can be integrated separately. Integration of these terms yields standard results such as:

• \( \int \frac{1}{x} \, dx = \ln|x| + C \)
• \( \int -\frac{1}{x^2} \, dx = \frac{1}{x} + C \)
• \( \int \frac{1}{x-1} \, dx = \ln|x-1| + C \)

Each term contributes to the final integrated expression, leading to a solution that captures the entire behavior of the original function.

Combining these successfully calculated integrals gives us the answer: \( \ln|x| - \frac{1}{x} + \ln|x-1| + C \). This result represents the accumulated value from the original rational expression over the given range.