Integration by Partial Fractions
Integration by Partial Fractions: We know that a rational function is a ratio of two polynomials P(x)/Q(x), where Q(x) ≠ 0. Now, if the degree of P(x) is lesser than the degree of Q(x), then it is a proper fraction, else it is an improper fraction. Even if a fraction is improper, it can be reduced to a proper fraction by the long division process.
So, if P(x)/Q(x) is an improper fraction, then P(x)/Q(x) = T(x) + P1(x)/Q(x) … where T(x) is a polynomial and P1(x)/Q(x) is a proper rational fraction. We already know how to integrate polynomials and in this article, we will be learning about integration by partial fractions. Also, the rational functions that we will consider are those whose denominators can be factorised into linear and quadratic equations.
Different Forms Integration by Partial Fractions
Let’s say that we want to evaluate ∫ [P(x)/Q(x)] dx, where P(x)/Q(x) is a proper rational fraction. In such cases, it is possible to write the integrand as a sum of simpler rational functions by using partial fraction decomposition. Post this, integration can be carried out easily. The following image indicates some simple partial fractions which can be associated with various rational functions:
Please note that A, B, and C are real numbers and their values should be determined suitably.
Some Examples of Integration by Partial Fractions
Question 1: Find ∫ dx / [(x + 1) (x + 2)]
Answer : The integrand is a proper rational function. Therefore, by using the form of partial fraction from the image above, we have:
1 / [(x + 1) (x + 2)] = A / (x + 1) + B / (x + 2) … (1)
Solving this equation, we get,
A (x + 2) + B (x + 1) = 1
Or, Ax + 2A + Bx + B = 1
x (A + B) + (2A + B) = 1
For LHS to be equal to RHS, we have
A + B = 0 and 2A + B = 1. On solving these two equations, we get
A = 1 and B = – 1.
Therefore, we have
1 / [(x + 1) (x + 2)] = 1 / (x + 1) – 1 / (x + 2)
Hence, ∫ dx / [(x + 1) (x + 2)] = ∫ dx / (x + 1) – ∫ dx / (x + 2)
= log |x + 1| – log |x + 2| + C
Note: Equation (1) is true for all permissible values of x. Some authors use the symbol ‘≡’ to indicate that the statement is an identity and use the symbol ‘=’ to indicate that the statement is an equation, i.e., to indicate that the statement is true only for certain values of x.
Question 2: Find ∫ [(x2 + 1) / (x2 – 5x + 6)] dx
Answer : In this case, the integrand is NOT a proper rational function. Hence we divide (x2 + 1) by (x2 – 5x + 6) and get,
(x2 + 1) / (x2 – 5x + 6) = 1 + (5x – 5) / (x2 – 5x + 6)
= 1 + (5x – 5) / (x – 2) (x – 3)
Now, let’s look at the second half of the above equation and let
(5x – 5) / (x – 2) (x – 3) = A / (x – 2) + B / (x – 3)
On solving it, we get
5x – 5 = A (x – 3) + B (x – 2) = Ax – 3A + Bx – 2B = x (A + B) – (3A + 2B)
Comparing the coefficients of the x term and constants, we get
A + B = 5 and 3A + 2B = 5. Further, on solving these two equations, we get
A = – 5 and B = 10.
Hence, we have
(x2 + 1) / (x2 – 5x + 6) = 1 – 5 / (x – 2) + 10 / (x – 3)
Therefore, ∫ [(x2 + 1) / (x2 – 5x + 6)] dx = ∫ dx – 5 ∫ 1 / (x – 2) + 10 ∫ 1 / (x – 3)
= x – 5log |x – 2| + 10log |x – 3| + C
Question 3: Find ∫ [(3x – 2) / (x + 1)2 (x + 3)] dx
Answer : If you look at the image above, then the denominator in this case is similar to example 4. Hence, we have
(3x – 2) / (x + 1)2 (x + 3) = A / (x + 1) + B / (x + 1)2 + C / (x + 3)
On solving it, we get
3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1) = Ax2 + 4Ax + 3A + Bx + 3B + Cx2 + 2Cx + C
= x2 (A + C) + x (4A + B + 2C) + (3A + 3B + C)
Comparing the coefficients of x2, x and the constant terms, we get
A + C = 0
4A + B + 2C = 3
3A + 3B + C = – 2
On solving these three equations, we get
A = 11/4, B = –5/2 and C = –11/4
Hence, we have
(3x – 2) / (x + 1)2 (x + 3) = 11 / 4(x + 1) – 5 / 2(x + 1)2 – 11 / 4(x + 3)
Therefore, ∫ [(3x – 2) / (x + 1)2 (x + 3)] dx = 11/4 ∫ dx / (x + 1) – 5/2 ∫ dx / (x + 1)2 – 11/4 ∫ dx / (x + 3)
= 11/4 log |x + 1| + 5 / 2(x + 1) – 11/4 log |x + 3| + C
= 11/4 log |(x + 1) / (x + 3)| + 5 / 2(x + 1) + C
Example 4: Find ∫ [(x2 + x + 1) / (x + 2) (x2 + 1)] dx
Solution: Now look at the example 5 in the image above. This example is similar to that example. Hence, we have
(x2 + x + 1) / (x + 2) (x2 + 1) = A / (x + 2) + (Bx + C) / (x2 + 1)
On solving this equation, we get
x2 + x + 1 = A (x2 + 1) + (Bx + C) (x + 2) = Ax2 + A + Bx2 + 2Bx + Cx + 2C
= x2 (A + B) + x (2B + C) + (A + 2C)
Comparing the coefficients of x2, x and the constant terms, we get
A + B = 1
2B + C = 1
A + 2C = 1
On solving these three equations, we get
A = 3/5, B = 2/5 and C = 1/5
Hence, we have
(x2 + x + 1) / (x + 2) (x2 + 1) = 3 / 5(x + 2) + [(2/5) x + 1/5] / (x2 + 1) = 3 / 5 (x + 2) + 1/5 [(2x + 1) / (x2 + 1)]
Therefore, ∫ [(x2 + x + 1) / (x + 2) (x2 + 1)] dx = 3/5 ∫ dx / (x + 2) + 1/5 ∫ [2x / (x2 + 1)] dx + 1/5 ∫ [1 / (x2 + 1)] dx
= 3/5 log |x + 2| + 1/5 log |x2 + 1| + 1/5 tan–1 x + C
More Solved Examples :
Question: Find ∫ [(3x – 1) / (x – 1) (x – 2) (x – 3)] dx
Solution: The integrand can be written as follows:
(3x – 1) / (x – 1) (x – 2) (x – 3) = A / (x – 1) + B / (x – 2) + C / (x – 3)
Multiplying both sides by {(x – 1) (x – 2) (x – 3)}, we get
3x – 1 = A (x – 2) (x – 3) + B (x – 1) (x – 3) + C (x – 1) (x – 2) … (a)
Now, let’s substitute x = 1 in equation (a). Hence, we have
3 – 1 = A (1 – 2) (1 – 3) + B (1 – 1) (1 – 3) + C (1 – 1) (1 – 2)
Or, 2 = A (–1) (–2) + B (0) (–2) + C (0) (–1) = 2A
Therefore, we get A = 1
Next, let’s substitute x = 2 in equation (a). Hence, we get
6 – 1 = {A (2 – 2) (2 – 3) + B (2 – 1) (2 – 3) + C (2 – 1) (2 – 2)}
Or, 5 = A (0) (–1) + B (1) (¬–1) + C (1) (0) = –B
Therefore, we get B = –5
Finally, let’s substitute x = 3 in equation (a). Hence, we get
9 – 1 = {A (3 – 2) (3 – 3) + B (3 – 1) (3 – 3) + C (3 – 1) (3 – 2)}
Or, 8 = A (1) (0) + B (2) (0) + C (2) (1) = 2C
Therefore, we get C = 4.
Hence, we now have
(3x – 1) / (x – 1) (x – 2) (x – 3) = 1 / (x – 1) – 5 / (x – 2) + 4 / (x – 3)
Therefore,
∫ [(3x – 1) / (x – 1) (x – 2) (x – 3)] dx = ∫ [1 / (x – 1)] dx – 5 ∫ [1 / (x – 2)] dx + 4 ∫ [1 / (x – 3)] dx
= log |x – 1| – 5log |x – 2| + 4 log |x – 3| + C
Solved Question :
Question 1: Why do we use partial fractions?
Answer: We decompose fractions into partial fractions like this as it makes specific integrals much simpler to do, and. We use it in the Laplace transform, which we meet later.
Question 2: How do you know when to use partial fractions in integration?
Answer: We can only do partial fractions if the degree of the numerator is severely less than the degree of the denominator. This is an important point. Thus, once you have understood that partial fractions can be done, you will need to factor the denominator as completely as possible
Question 3: How do you integrate fractions?
Answer: If someone asks you to integrate a fraction, you must try to multiply or divide the top and bottom of the fraction by a number. Occasionally it will be of help if you split a fraction up prior to making an attempt to integrate it. You can use the method of partial fractions for this.
Question 4: What is mean integration?
Answer: Integration happens when we bring distinct people or things together. For instance, the integration of students from all of the city’s colleges at the university and more. We have all come across the word differentiate that means to “set apart.” Thus, integrate is the opposite of this.
Integral Calculus Formulas
Integration is the reverse process of differentiation. So, we may call it as Inverse Differentiation. Integration is the process to find a function with its derivative. Basic integration formulas on different functions are very useful and important. This article deals with the concept of integral calculus formulas with concepts and examples. Integral calculus is the branch of mathematics dealing with the formulas for integration, and classification of integral formulas. The student will take benefits from this concrete article. Let us learn the concept and the integral calculus formulas.
Concept of the Integral Calculus
Integration is the algebraic method to find the integral for a function at any point on the graph. The integral comes from not only to determine the inverse process of taking the Derivative.
But also for solving the area problem as well. The integral of the function of x from range a to b will be the sum of the rectangles to the curve at each interval of change in x as the number of rectangles goes to infinity.
Given a function, f(x), is an anti-derivative of f(x), is any function F(x) such that
F ′ (x)=f(x)
If F(x) is any anti-derivative of f(x), then the most general anti-derivative of f(x) will be its indefinite integral.
The integral of a function f(x) with respect to variable x is:
∫f(x)dx
Also, integration is considered as almost an inverse to the operation of differentiation means that if,
The extra C called the constant of integration, which is really necessary. Definite integrals are the special kind of integration, where both endpoints are fixed. So, it always represents some bounded region, for computation.
Some Properties
∫kf(x)dx=k∫f(x)dx where k is any number. So, we factor the multiplicative constants out of indefinite integrals.
∫–f(x)dx=–∫f(x)dx This is really the first property with k=−1k=−1 and so no proof of this property will be given.
∫f(x)±g(x)dx=∫f(x) dx±∫g(x)dx It means the integral of a sum or difference of functions is the sum or difference of their individual integrals. This is applicable and useful for many functions as we need.
Formula for Integral Calculus
Solved Examples for Integral Calculus Formulas
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