Problem 21 - Calculus Single Variable

Question

In each of Exercises \(21-30\), use the method of partial fractions to decompose the integrand. Then evaluate the given integral. \(\int \frac{3 x^{2}-5 x+4}{(x-1)\left(x^{2}+1\right)} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(2\ln|x-1| + \frac{1}{2}\ln(x^2+1) - 4\tan^{-1}(x) + C\).

Step 1: Identify the Form of Partial Fraction Decomposition

To decompose the fraction \( \frac{3x^2 - 5x + 4}{(x-1)(x^2 + 1)} \), note that the denominator is a product of linear term \((x-1)\) and an irreducible quadratic term \((x^2 + 1)\). Hence, the partial fraction decomposition is of the form: \[ \frac{A}{x-1} + \frac{Bx + C}{x^2 + 1} \] where \(A, B,\) and \(C\) are constants to be determined.

Step 2: Set Up the Equation for Constants

To find the constants \(A, B,\) and \(C\), equate the numerators after multiplying through by the common denominator:\[ 3x^2 - 5x + 4 = A(x^2 + 1) + (Bx + C)(x - 1) \] Simplify the right side to \[ A(x^2 + 1) + (Bx + C)(x - 1) = Ax^2 + A + Bx^2 - Bx + Cx - C \] \[ = (A + B)x^2 + (-B + C)x + (A - C) \] Set this equal to the numerator \(3x^2 - 5x + 4\) and compare coefficients.

Step 3: Solve for Coefficients A, B, C

Compare the coefficients:1. Coefficient of \(x^2\): \( A + B = 3 \) 2. Coefficient of \(x\): \( -B + C = -5 \) 3. Constant term: \( A - C = 4 \) Solve these equations:- From (1): \( A = 3 - B \)- Substitute into (3): \( 3 - B - C = 4 \rightarrow -B - C = 1\)- Solve the system: \( A = 2, \ B = 1, \ C = -4 \).

Step 4: Write the Integral in Partial Fractions

Substitute \(A, B,\) and \(C\) back into the decomposition:\[ \frac{3x^2 - 5x + 4}{(x-1)(x^2 + 1)} = \frac{2}{x-1} + \frac{x - 4}{x^2 + 1} \] Now the integral becomes two separate integrals:\[ \int \left( \frac{2}{x-1} + \frac{x - 4}{x^2 + 1} \right) dx \] This splits into:\[ \int \frac{2}{x-1} dx + \int \frac{x - 4}{x^2 + 1} dx \]

Step 5: Evaluate the Integral of the First Term

The integral \( \int \frac{2}{x-1} dx \) is a simple logarithmic form:\[ 2 \ln |x-1| + C_1 \]

Step 6: Evaluate the Integral of the Second Term

For \( \int \frac{x}{x^2 + 1} dx \), use substitution. Let \( u = x^2 + 1 \), so \( du = 2x \, dx \) and \( \frac{1}{2}du = x\, dx \). This simplifies to:\[ \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| = \frac{1}{2} \ln(x^2 + 1) \] For \( \int \frac{-4}{x^2 + 1} dx \), note this is \(-4\) times the integral of \( \frac{1}{x^2 + 1} dx \), which is:\[ -4 \tan^{-1}(x) + C_2 \]

Step 7: Combine the Integrated Results

Combine and simplify the results of all calculated integrals:\[ \int \frac{3x^2 - 5x + 4}{(x-1)(x^2 + 1)} dx = 2 \ln|x-1| + \frac{1}{2} \ln(x^2 + 1) - 4 \tan^{-1}(x) + C \] where \(C\) represents the constant of integration.

Key Concepts

Integrals

Integration is a key concept in calculus that involves finding a function whose derivative matches a given function. Imagine integration as the reverse process of differentiation. It helps in determining the area under a curve, among other applications. The integral in our exercise is:
\[ \int \frac{3x^2 - 5x + 4}{(x-1)(x^2 + 1)} \, dx \]
This is a rational function where the degree of the numerator is less than the degree of the denominator. Hence, it can be decomposed using partial fractions, making the integration more manageable. Partial fraction decomposition is particularly helpful when dealing with complex rational functions that aren’t straightforward to integrate directly.

Logarithmic Integration

Logarithmic integration involves integrating functions that convert into logarithmic forms. These types often include simple rational expressions like \( \frac{1}{x} \).
In our case, we encounter the integral \( \int \frac{2}{x-1} \, dx \).
This integral can be evaluated using a direct application of the natural logarithm rule:

• The indefinite integral of \( \frac{1}{x-a} \) is \( \ln |x-a| + C\).
• Applying this, \( \int \frac{2}{x-1} \, dx = 2 \ln |x-1| + C_1 \).

The constant value, \(C_1\), represents the constant of integration that arises whenever working with indefinite integrals.

Arctangent Integration

Arctangent integration is often involved when the integrand has the form \( \frac{1}{x^2 + 1} \).
This matches the derivative of the \( \tan^{-1}(x) \) function.

• The basic form of this integral is \( \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) + C\).

In our problem, we find the term:\[ \int \frac{-4}{x^2 + 1} \, dx \]Since \( -4 \, \int \frac{1}{x^2 + 1} \, dx \) gives \( -4 \tan^{-1}(x) + C_2 \), this reflects the integration of arctangent multiplied by a constant factor. It's a popular form that surfaces when handling rational functions with quadratic terms.

Algebraic Manipulation

Algebraic manipulation plays a critical role in solving integrals, especially when utilizing partial fraction decomposition.
This process involves breaking down a complex rational expression into simpler parts:
1. Consider \( \frac{3x^2 - 5x + 4}{(x-1)(x^2 + 1)} \), where the decomposition gives \( \frac{A}{x-1} + \frac{Bx + C}{x^2 + 1} \).2. By equating coefficients and solving systems of equations for \( A, B, \) and \( C \), we get simple fractions:

• Solve \( A + B = 3 \),
• \( -B + C = -5 \),
• \( A - C = 4 \).

3. Once simplified, these fractions can be integrated separately. This crucial step allows us to handle each part with the most applicable integration technique, including logarithmic and arctangent methods.