Math: Partial Fractions and Integration
Decompose rational functions and use them to evaluate integrals
Math: Partial Fractions and Integration
Decompose rational functions and use them to evaluate integrals
Math - Grade 9-12
- 1
Decompose the rational expression into partial fractions: 5/(x(x + 2)).
Start with A/x + B/(x + 2), then multiply both sides by x(x + 2).
The decomposition is 5/(x(x + 2)) = 5/(2x) - 5/(2(x + 2)). This is found by writing 5/(x(x + 2)) = A/x + B/(x + 2), so 5 = A(x + 2) + Bx, which gives A = 5/2 and B = -5/2. - 2
Use partial fractions to evaluate the indefinite integral: integral of 5/(x(x + 2)) dx.
The integral is (5/2)ln|x| - (5/2)ln|x + 2| + C. This follows from the decomposition 5/(x(x + 2)) = 5/(2x) - 5/(2(x + 2)). - 3
Decompose the rational expression into partial fractions: (3x + 7)/((x - 1)(x + 4)).
Substitute x = 1 and x = -4 after clearing the denominator.
The decomposition is (3x + 7)/((x - 1)(x + 4)) = 2/(x - 1) + 1/(x + 4). After multiplying by (x - 1)(x + 4), we get 3x + 7 = A(x + 4) + B(x - 1), and solving gives A = 2 and B = 1. - 4
Evaluate the indefinite integral: integral of (3x + 7)/((x - 1)(x + 4)) dx.
The integral is 2ln|x - 1| + ln|x + 4| + C. This comes from the decomposition (3x + 7)/((x - 1)(x + 4)) = 2/(x - 1) + 1/(x + 4). - 5
Decompose the rational expression into partial fractions: (2x + 1)/(x^2 - 9).
First factor x^2 - 9 as a difference of squares.
The decomposition is (2x + 1)/(x^2 - 9) = 7/(6(x - 3)) + 5/(6(x + 3)). Since x^2 - 9 = (x - 3)(x + 3), we write 2x + 1 = A(x + 3) + B(x - 3), which gives A = 7/6 and B = 5/6. - 6
Evaluate the indefinite integral: integral of (2x + 1)/(x^2 - 9) dx.
The integral is (7/6)ln|x - 3| + (5/6)ln|x + 3| + C. This result uses the partial fraction form 7/(6(x - 3)) + 5/(6(x + 3)). - 7
Decompose the rational expression into partial fractions: (x + 5)/(x^2 + 5x + 6).
Factor the quadratic denominator before setting up the decomposition.
The decomposition is (x + 5)/(x^2 + 5x + 6) = 3/(x + 2) - 2/(x + 3). Since x^2 + 5x + 6 = (x + 2)(x + 3), we write x + 5 = A(x + 3) + B(x + 2), which gives A = 3 and B = -2. - 8
Evaluate the indefinite integral: integral of (x + 5)/(x^2 + 5x + 6) dx.
The integral is 3ln|x + 2| - 2ln|x + 3| + C. This follows by integrating the decomposition 3/(x + 2) - 2/(x + 3). - 9
Decompose the rational expression with a repeated factor: (4x + 1)/(x(x + 1)^2).
For the repeated factor (x + 1)^2, include both B/(x + 1) and C/(x + 1)^2.
The decomposition is (4x + 1)/(x(x + 1)^2) = 1/x - 1/(x + 1) + 4/(x + 1)^2. Multiplying by x(x + 1)^2 gives 4x + 1 = A(x + 1)^2 + Bx(x + 1) + Cx, and solving gives A = 1, B = -1, and C = 4. - 10
Evaluate the indefinite integral: integral of (4x + 1)/(x(x + 1)^2) dx.
The integral is ln|x| - ln|x + 1| - 4/(x + 1) + C. This comes from integrating 1/x - 1/(x + 1) + 4/(x + 1)^2 term by term. - 11
Evaluate the definite integral from x = 2 to x = 4 of 6/(x^2 - 1) dx.
Factor x^2 - 1 and use the limits after you find an antiderivative.
The value of the definite integral is 3ln(5/3). Since 6/(x^2 - 1) = 3/(x - 1) - 3/(x + 1), an antiderivative is 3ln|x - 1| - 3ln|x + 1|. Evaluating from 2 to 4 gives 3ln(3/5) - 3ln(1/3), which simplifies to 3ln(9/5) if calculated from that decomposition. However, the correct decomposition is 3/(x - 1) - 3/(x + 1), so the evaluation is 3ln(3) - 3ln(5) - [3ln(1) - 3ln(3)] = 3ln(9/5). - 12
The rational function (x^2 + 3x + 1)/(x + 1) is improper because the numerator has degree 2 and the denominator has degree 1. Rewrite it using polynomial division, then integrate.
Divide first. Partial fractions are used only after the rational expression is proper.
Polynomial division gives (x^2 + 3x + 1)/(x + 1) = x + 2 - 1/(x + 1). Therefore, the integral is x^2/2 + 2x - ln|x + 1| + C.