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Integration by Partial Fractions is a method of integration used for complex functions. It is used to decompose and then integrate a rational fraction integrand having complex terms in its denominator. Integration is a method used to define and calculate the area of the region bounded by the graph of functions.
- Partial Fractions are defined as the sum of proper rational functions obtained on decomposing an improper rational function.
- Improper Rational Function is a rational function whose degree of the numerator is not less than the degree of the denominator.
- The complex rational expressions cannot be solved in a simpler way.
- Thus, partial fractions are used to decompose the rational expressions into simpler partial fractions.
Integration by Partial Fractions involves decomposing the improper rational function into a proper rational function and then integrating the same. It is given as:
| ∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx |
Where, f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and g(x) = q(x).s(x).
Key Terms: Integration by Partial Fractions, Partial Fractions, Rational Functions, Fraction, Numerator, Denominator, Integration
What is Integration by Partial Fractions?
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Integration by Partial Fractions is one of the three methods of integration with Integration by Substitution and Integration by Parts.
- In integration by partial fractions, the partial rational fraction is decomposed into a sum of simpler rational fractions.
- Rational Functions are considered integrands in the process of integration.
- Their denominators can be factored into linear and quadratic factors.
- The rational fraction is decomposed into simpler rational fractions through the process of Partial Fraction Decomposition.
- Partial Fraction Decomposition is the process of writing the integrand as a sum of simpler rational functions.
Introduction to Integrals Detailed Video Explanation
Example: Consider the Partial Fraction Decomposition of 2/(x+1) - 1/x.
Adding them up, we get
2/(x+1) - 1/x = (x-1)/(x2+x)
Now, (x-1)/(x2+x) can be decomposed into
(x-1)/(x2+x) = 2/(x+1) - 1/x
The given partial fractions are now decomposed into simpler terms and can be integrated easily.
Integration by Partial Fractions would now be done as:
| ∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx |
Where
- f(x)/g(x) = p(x)/q(x) + r(x)/s(x)
- g(x) = q(x).s(x)
Integration by Partial Fractions Formula
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Integration by Partial Fractions involves the decomposition of partial fractions to proper rational fractions. Given below are the formulas that are used to decompose the given improper rational functions. They are used to write down the integrand as a sum of proper rational functions easily.
| Rational Fractions | Partial Fractions |
|---|---|
| (px + q) / (x-a) (x – b), where a ≠b | A / (x – a) + B / (x-b) |
| (px + q) / (x-a)2 | A / (x – a) + B / (x-a)2 |
| (px2 + qx + r) / (x-a) (x-b) (x-c) | A / (x – a) + B/ (x-b) + C/ (x-c) |
| (px2 + qx + r) / (x – a)2 (x - b) | A / (x-a) + B/ (x-a)2 + C/ (x-b) |
| (px2 + qx + r) / (x2 – bx + c) | A / (x-a) + (Bx + C) / (x2 – bx + c) |
| Where x2 – bx + c cannot be factorized further. | |
Read More: NCERT Solutions For Class 12 Mathematics Integrals
How to do Integration by Partial Fractions?
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Integration by Partial Fraction can be carried out using the given steps as follows:
- Step 1: First confirm whether the given integrand is a proper or improper rational function.
- Step 2: In case of an improper rational function, identify the type of denominator.
- Step 3: Now, decompose the integrand with the help of a suitable form from the five different forms mentioned above.
- Step 4: Lastly, divide the integration into parts and integrate the individual functions.
Solved ExampleExample: Evaluate ∫[6/(x2-1)]dx using Integration by Partial Fraction. Solution: Using algebraic identities, it is known that x2-1 = (x+1)(x-1). Thus, ∫[6/(x2-1)]dx can be written as ∫[6/(x+1)(x-1)]dx. Using the suitable form of partial fraction for the given rational form, we get 6/(x+1)(x-1) = A/(x-1) + B/(x+1) In order to make a common denominator on both sides, find the value of A and B.
The denominators are equal on both sides, thus, the numerators will also be equal. 6 = [A(x+1) + B (x-1)] On simplification,
Thus, 6/(x+1)(x-1) = 3/(x-1) + (-3)/(x+1) ∫[6/(x2-1)]dx = ∫[3/(x-1) - 3/(x+1)]dx On solving the same, we get ∫[6/(x2-1)]dx = −3ln(|x+1|)+3ln(|x−1|) + C |
Integration by Partial Fractions Examples
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Here are some solved examples on the method of Integration by Partial Fractions.
Example 1: Integrate ∫[3x2 + x + 3]/[(x−1)3(x2+1)]dx using Integration by Partial Fractions.
Solution: Given expression is ∫[3x2 + x + 3]/[(x−1)3(x2+1)]dx.
First, perform the Partial Fraction Decomposition of the given expression.
[3x2+x+3]/[(x−1)3(x2+1)] = A/(x−1) + B/ (x−1)2 + C/(x−1)3 + (Dx+E)/(x2+1)
On cross-multiplication,
3x2+x+3 = A(x−1)2(x2+1) + B(x−1)(x2+1) + C(x2+1) + (Dx+E)(x−1)3
Substitute x = 1, so, C = 72 ...(1)
Comparing the coefficients of
- x4 ⇒ 0 = A + D. ..(2)
- x3 ⇒ 0 = −2A + B −3D + E ...(3)
- x2 ⇒ 3 = 2A−B+C+3D−3E ...(4)
- x ⇒ 1 = −2A + B − D + 3E ...(5)
Sum up (4) and (5) to obtain,
4 = C + 2D
D = 1/4 (From (1))
A = −1/4 (From (2))
Sum up (3) and (4) to obtain,
3 = C − 2E
E = 1/4 (Using (1))
Using (5), B = 0
Thus, the partial fraction is (−1/4)/(x−1) + (7/2)/(x−1)3 + [(1/4)x+(1/4)]/(x2+1).
Now, the integral will be
I = (−1/4)∫[1/(x−1)]dx + 7/2 ∫[1/(x−1)3]dx + 1/4∫[x/(x2+1)]dx + 1/4∫[1/(x2+1)]dx
I = −1/4 ln|x−1| − 7/4(x−1)2 + 1/8 ln(x2+1) + 1/4 tan−1x + C
I = (1/4) {tan−1x − 7/(x−1)2 + ln((√(x2+1))/(|x−1|))} + C
Thus, ∫[3x2 + x + 3]/[(x−1)3(x2+1)]dx = (1/4){tan−1x − 7/(x−1)2 + ln((√(x2+1))/(|x−1|))} + C.
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Example 2: What will be the value of \(\begin{array}{l}\int \frac{x}{(x – 1)(x – 2)(x – 3)}\ dx\end{array}\).
Solution: Given expression is \(\begin{array}{l}\int \frac{x}{(x – 1)(x – 2)(x – 3)}\ dx\end{array}\).
It is an improper rational function and needs to be decomposed into the sum of proper rational functions.
\(=\frac{x}{(x – 1)(x – 2)(x – 3)}= \frac{A}{(x – 1)}+\frac{B}{(x – 2)} + \frac{C}{(x – 3)}\\\)
\(= \frac{A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2)}{(x – 1)(x – 2)(x – 3)}\)
On equating the denominators, we get
- x = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2)
- x = A(x2 – 5x + 6) + B(x2 – 4x + 3) + C(x2 – 3x + 2)
- x = (A + B + C)x2 – (5A + 4B + 3C)x + (6A + 3B + 2C)
Now, equate the coefficients of the terms to obtain
- A + B + C = 0….(i)
- 5A + 4B + 3C = -1….(ii)
- 6A + 3B + 2C = 0….(iii)
Using (ii),
2A + B + 3(A + B + C) = -1
2A + B + 0 = -1 = -1….(iv)
Using (iii),
4A + B + 2(A + B + C) = 0
4A + B + 0 = 0
B = -4A….(v)
Using (iv) and (v),
2A – 4A = -1
-2A = -1
- A = ½
- B = -4(½) = -2
- C = -(½) + 2 = 3/2
Put the obtained values in the decomposed form of the integrand.
=\(\begin{array}{l}\frac{x}{(x – 1)(x – 2)(x – 3)}= \frac{1}{2(x – 1)}-\frac{2}{(x – 2)} + \frac{3}{2(x – 3)}\end{array}\)
= \(\begin{array}{l}\int \frac{x}{(x – 1)(x – 2)(x – 3)}\ dx= \frac{1}{2}\int \frac{1}{(x – 1)}\ dx – 2\int \frac{1}{(x – 2)}\ dx + \frac{3}{2} \int \frac{1}{(x – 3)}\ dx\end{array}\)
= (1/2) log|x – 1| – 2 log|x – 2| + (3/2) log|x – 3| + C
Thus, \(\begin{array}{l}\int \frac{x}{(x – 1)(x – 2)(x – 3)}\ dx\end{array}\) = (1/2) log|x – 1| – 2 log|x – 2| + (3/2) log|x – 3| + C.
Things to Remember
- Integration by Partial Fractions is used to decompose and integrate a rational fraction with complex terms in the denominator.
- Integration by Substitution, Integration by Partial Fractions, and Integration by Parts are three methods of integration.
- It involves the partial fraction decomposition of proper rational fractions into a sum of simpler rational fractions.
- Partial Fraction Decomposition is a method of expressing an improper rational function as a sum of simpler rational functions.
- Integration by Partial Fractions is given as ∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx.
- There are five different forms of partial fractions to decompose specific forms of proper rational fractions.
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Sample Questions
Ques. What are Partial Fractions? (3 Marks)
Ans. Partial Fractions are defined as the fractions which are used for the decomposition of a rational expression.
- An algebraic fraction can be broken down into a sum of two or more rational expressions known as “partial fractions“.
- They are the reverse or opposite of the addition of rational expressions.
- Partial fraction has a numerator and denominator, in which the denominator represents the decomposed part of a rational function.
Ques. What are the various types of denominators in Partial Fractions? (2 Marks)
Ans. In partial fractions, there are four distinct types of denominators:
- Linear Factors
- Irreducible Factors of Degree 2
- Repeated Linear Factors
- Repeated Irreducible Factors of Degree 2
Ques. Find ∫ dx / [(x + 1) (x + 2)]. (3 Marks)
Ans. Given expression is ∫ dx / [(x + 1) (x + 2)].
The integrand is a rational function in the proper sense. Thus, using the suitable form of partial factions, we get
1 / [(x + 1) (x + 2)] = A / (x + 1) + B / (x + 2) … (1)
On solving the equation,
A (x + 2) + B (x + 1) = 1
Or, Ax + 2A + Bx + B = 1
x (A + B) + (2A + B) = 1
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.
Ques. How is Integration by Partial Fractions performed? (1 Mark)
Ans. Integration by Partial Fractions is used to decompose and then integrate a rational fraction integrand which has complex terms in the denominator. Integration by Partial Fractions helps to calculate and decompose the expression into simpler terms so as to calculate or integrate the expression thus obtained.
Ques. What is Integration? (3 Marks)
Ans. Integration is a method used to find the area of the region under the curve. Integration is the inverse process of differentiation.
- It is defined as a process of finding the antiderivative of a function.
- Integration is used to find the area of the region bounded by the graph of functions.
- The notion of integration was proposed by Leibniz.
Ques. Find ∫ [(x2 + 1) / (x2 – 5x + 6)] dx. (5 Marks)
Ans. Given expression is ∫ [(x2 + 1) / (x2 – 5x + 6)] dx.
The integrand in this case is not a proper rational function. Thus, on decomposing the same:
(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)
On observing the second half of the equation, it can be stated that,
(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)
We get A + B = 5 and 3A + 2B = 5 by comparing the coefficients of the x term and constants. We also get A = – 5 and B = 10 when we solve these two equations.
As a result, 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.
Ques. When is Integration by Partial Fractions used? (2 Marks)
Ans. Integration by Partial Fractions is used to factor the denominator and then decompose them into two different fractions where the denominators are the factors respectively and the numerator is calculated suitably. Thus, Integration by Partial Fractions is used when the degree of the denominator is more than the numerator and the denominator is a complicated expression.
Ques. Find ∫ [(3x – 2) / (x + 1)2 (x + 3)] dx. (5 Marks)
Ans. Given that ∫ [(3x – 2) / (x + 1)2 (x + 3)] dx.
Using the suitable form of partial factions, we get
(3x – 2) / (x + 1)2 (x + 3) = A / (x + 1) + B / (x + 1)2 + C / (x + 3)
On solving the same, 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
By solving these equations, A = 11/4, B = –5/2, and C = –11/4.
Thus,
(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.
Ques. State the applications of Integration by Partial Fraction. (3 Marks)
Ans. The applications of Integration by Partial Fraction are:
- It is majorly used to decompose the fraction into two or more different fractions.
- It is used to integrate Rational Fractions in Calculus.
- It is used to find the Inverse Laplace Transform in the theory of differential equations.
Ques. What are the different Methods of Integration? (2 Marks)
Ans. There are three Methods of Integration which are as follows:
- Integration by Substitution Method
- Integration by Parts
- Integration by Partial Fractions
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