Question

Use the method of partial fractions to evaluate each of the following integrals. \(\int \frac{2-x}{x^{2}+x} d x\)

Short answer

The evaluated integral is \(2 \ln |x| - 3 \ln |x+1| + C\).

Step-by-step solution

Factor the Denominator

First, we need to factor the denominator of the integrand \(x^2 + x\). By factoring, we have \(x(x + 1)\). So the integral becomes:\[\int \frac{2-x}{x(x + 1)} \, dx.\]

Decompose into Partial Fractions

We express \(\frac{2-x}{x(x + 1)}\) in terms of partial fractions. This means writing it in the form:\[\frac{2-x}{x(x + 1)} = \frac{A}{x} + \frac{B}{x+1}\]where \(A\) and \(B\) are constants to be determined.

Solve for Constants A and B

Multiply both sides by the common denominator \(x(x + 1)\) to eliminate the fractions:\[2 - x = A(x + 1) + Bx.\]Expanding the right side gives:\[2 - x = Ax + A + Bx.\]Combine like terms:\[2 - x = (A + B)x + A.\]Equate coefficients:1. \(A + B = -1\)2. \(A = 2\)Solve these equations: From \(A = 2\), substitute into \(A + B = -1\) giving \(2 + B = -1\), hence \(B = -3\). So, \(A = 2\) and \(B = -3\).

Rewrite the Integral Using Partial Fractions

Substitute the values of \(A\) and \(B\) back into the decomposition:\[\int \frac{2}{x} \, dx - \int \frac{3}{x + 1} \, dx.\]This separates the integral into two simpler integrals.

Integrate Each Term Separately

Integrate each term individually:1. \(\int \frac{2}{x} \, dx = 2 \ln |x| + C_1\), where \(C_1\) is a constant of integration.2. \(\int \frac{-3}{x + 1} \, dx = -3 \ln |x+1| + C_2\), where \(C_2\) is another constant.Combine the integrals: \[2 \ln |x| - 3 \ln |x+1| + C,\]where \(C\) is the overall constant of integration.

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a powerful algebraic technique used in calculus to simplify the integration of rational functions. By breaking down a complex fraction into simpler pieces, it becomes easier to integrate.

Here's how you can do it:

• Start with a fraction where the degree of the numerator is less than the degree of the denominator.
• Factorize the denominator into simpler linear factors, whenever possible.
• Then, express the original fraction as a sum of fractions whose denominators are these factors. Each fraction in the sum will have a constant numerator.

For instance, consider the task of decomposing \( \frac{2-x}{x(x + 1)} \). First, identify the potential partial fractions as \( \frac{A}{x} \) and \( \frac{B}{x+1} \).
To determine the constants \( A \) and \( B \), you work by clearing fractions and equating coefficients. This means multiplying both sides by the common denominator and comparing each term, effectively solving a system of linear equations.

Once the values for \( A \) and \( B \) are found, you can substitute them back into the fractional expression to simplify the integration process.

Integral Calculus

Integral calculus is central to finding areas under curves, solving differential equations, and in many physical and geometrical applications. An integral like \( \int \frac{2-x}{x^2+x} \, dx \) can be computed by simplifying the expression using techniques such as partial fraction decomposition.

Fundamental concepts in integration include:

• The integrand: The function to be integrated.
• The integral sign \( \int \): Represents the operation of integration.
• Differential \( dx \): Indicates the variable of integration, which is \( x \) in this case.
• Constant of Integration \( C \): Represents the family of antiderivatives.

After decomposing into partial fractions, the integral problem becomes manageable since each simplified term can be integrated separately. Integral calculus not only helps in solving such problems but also in understanding how quantities accumulate over time or space.

Logarithmic Integration

Logarithmic integration arises naturally when integrating functions of the form \( \frac{1}{x} \). This occurs frequently when the decomposition of fractions leads to terms like \( \frac{A}{x} \) or \( \frac{B}{x+1} \).

To integrate such terms, a key rule is:

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

The absolute value ensures correctness since the logarithm is only defined for positive numbers. When solving an integral such as \( \int \frac{2}{x} \, dx - \int \frac{3}{x+1} \, dx \), each fraction produces a logarithmic result:

• \( 2 \ln |x| \) comes from \( \int \frac{2}{x} \, dx \).
• \( -3 \ln |x+1| \) results from \( \int \frac{-3}{x+1} \, dx \).

Once computed, these terms can be combined to form the final expression of the integral, capturing how change accumulates across the given domain as logarithmic functions.