JEE Exam  >  JEE Notes  >  Mathematics (Maths) Class 12  >  Integration by Partial Fractions

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE PDF Download

Introduction

We know that a Rational Number can be expressed in the form of p/q, where p and q are integers and q≠0. Similarly, a rational function is defined as the ratio of two polynomials which can be expressed in the form of partial fractions: P(x)/Q(x), where Q(x)≠0. 

  1. Proper partial fraction: When the degree of the numerator is less than the degree of the denominator, then the partial fraction is known as a proper partial fraction.
  2. Improper partial fraction: When the degree of the numerator is greater than the degree of denominator then the partial fraction is known as an improper partial fraction. Thus, the fraction can be simplified into simpler partial fractions, that can be easily integrated.

In this section, we show how to integer any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions 2/(x – 1) and 1/(x + 2) to a common denominator we obtain  

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
If we now reverse the procedure, we see how to integrate the function on the right side of this equation   Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
To see how the method of partial fractions works in general, let's consider a rational function  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
Where P and Q are polynomials. It's possible to express f as sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper. Recall that if P(x) = Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
where an ≠ 0, then the degree of P is n and we write deg (P) = n.
If f is improper, that is, deg(P) Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE deg (Q), then we must take the preliminary step of dividing Q into P  (by division) until a remainder R(x) is obtained such that deg (R) < deg(Q). The division statement is

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE    where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required.


Ex.1 Evaluate Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEEdx.


Sol. Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write
Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

The next step is to factor the denominator Q(x) as far as possible . It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors

(of the form ax2 + bx + c, where b2 – 4ac < 0).

For instance, if Q(x) = x4 – 16, we could factor it as Q(x) = (x2 – 4) (x2 + 4) = (x – 2) (x + 2) (x2 + 4)
The third step is to express the proper rational function R(x)/Q(x) (from equation 1) as a sum of partial fractions of the form

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur.

Case I : The Denominator Q(x) is a product of distinct linear factors.

This means that we can write Q(x) = (a1x + b1) (a2x + b2) ... (akx + bk)
where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A1,A2,...,Ak such that.
Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
These constants can be determined as in the following example.


Ex.2 Evaluate  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEEdx.


Sol. Since the degree of the numerator is less than the degree of the denominator, we don't need to divide.

We factor the denominator as 2x3 + 3x2 – 2x = x(2x2 + 3x – 2) = x(2x – 1) (x + 2)

Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
To denominator the values of A, B and C, we multiply both sides of this equation by the product of the denominators, x(2x – 1) (x + 2), obtaining.

(4) x2 + 2x – 1 = A(2x – 1) (x + 2) + Bx(x + 2) + Cx(2x – 1)

Expanding the right side of equation 4 and writing it in the standard form for polynomials, we get


(5) x2 + 2x – 1 = (2A + B + 2C) x2 + (3A + 2B – C) x – 2A

The polynomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of xon the right side, 2A + B + 2C, must equal the coefficient of x2 on the left side-namely, 1. Likewise. The coefficients of x are equal and the constant terms are equal. This gives the following system of equation for A, B and C.

2A + B + 2C = 1  , 3A + 2B – C = 2  , – 2A = –1

Solving, we get  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

In integrating the middle term we have made the mental substitution u = 2x – 1, which gives du = 2dx and dx = du/2.

Case II : Q(x) is a product of linear factors, some of which are repeated

Suppose the first linear factor (a1x + b1) is repeated r times, that is, (a1x + b1)r occurs in the factorization of Q(x). Then instead of the single term A1/(a1x + b1) in equation 2, we would use

(6) Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

By way of illustration, we could write  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE


but we prefer to work out in detail a simpler example.


Ex.3 Evaluate  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEEdx

Sol.

The first step is to divide. The result of long division is  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

The second step is to factor the denominator Q(x) = x3 – x2 – x + 1. Since Q(1) = 0, we know that x – 1 is a factor and we obtain x3 – x2 – x + 1 = (x – 1) (x– 1) = (x – 1) (x – 1)(x + 1) = (x – 1)2 (x + 1) Since the linear factor x – 1 occurs twice, the partial fraction decompositoin is

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Multiplying by the least common denominator (x – 1)2 (x + 1), we get

(7) 4x = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)2 = (A + C) x2 + (B – 2x) x + (–A + B + C)

Now we equate coefficients : A + C = 0,B – 2C = 4, –A + B + C = 0

Solving, we obtain A = 1, B = 2, and C = – 1, so

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
 

Case III : Q(x) contains irreducible quadratic factors, none of which is repeated

If Q(x) has the factor ax2 + bx + c, where b2 – 4ac < 0, then in addition to the partial fractions in equation 2 and 6, the expression or R(x)/Q(x) will have a term of the form.

(8) Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE where A and B are constants to be determined. For instance, the function givenby f(x) = x/[(x – 2)(x2 + 1) (x2 + 4)] has a partial fraction decomposition of the form

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

The term given in (8) can be integrated by completing the square and using the formula.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE


Ex.4 Evaluate  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEEdx
Sol.

Since x3 + 4x = x(x2 + 4) can't be factored further, we write  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Multiplying by x(x2 + 4), we have 2x2 – x + 4 = A (x2 + 4) + (Bx + C) x = (A + B) x2 + Cx + 4A Equating coefficients, we obtain A + B =2 C = –1 4A = 4

Thus A = 1, B = 1 and C = –1 and   Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

In order to integrate the second term we split it into to parts   Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

We make the substitution u = x2 + 4 in the first of these integrals so that du = 2x dx. We evaluate the second integral by means of Formula 9 with a = 2.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Case IV : Q(x) Contains A repeated irreducible quadratic factor.

If Q (x) has the factor (ax2 + bx + c)r, where b2 – 4ac < 0, then instead of the single partial fraction
Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
occurs in the partial fraction decomposition of R(x)/Q(x), each of the terms in (10) can be integrated by first completing the square.


Ex.5 Write out the form of the partial fraction decomposition of the function

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE
 
Sol.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

 

Ex.6 Evaluate  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Sol. The form of the partial fraction decomposition is  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Multiplying by x(x2 + 1)2, we have –x3 + 2x– x + 1 = A(x2 + 1)2 + (Bx + C) x (x2 + 1) + (Dx + E)x

= A(x+ 2x2 + 1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex = (A + B) x4 + Cx3 + (2A + B + D) x2 + (C + E) x + A

If we equate coefficient, we get the system    A + B = 0 ,C = –1  ,2A + B + D = 2  ,C+E =–1 , A=1

Which the solution A = 1, B = –1, D = 1, and E = 0. Thus

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

could be evaluated by the method of case III, it's much easier to observe that if u = x(x+ 3) = x3 + 3x, then du = (3x+ 3) dx and so  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Ex.7 EvaluateIntegration by Partial Fractions | Mathematics (Maths) Class 12 - JEEdx


Sol. In this example there is a repeated quadratic polynomial in the denominator. Hence, according to our previous discussion

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

For some constants A1, B1, A2 and B2
An easy way to determine these constant is as follows. By long division,

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Thus A= 1, B1 = – 3, A2 = 1 and B= 0

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE


Ex.8 Evaluate  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Sol.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

put sin x – cos x = s and sin x + cosx = t ⇒ (cosx + sinx) dx = ds and (cosx – sin x) dx = dt

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Ex.9 Evaluate  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Sol.

We have  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE , [multiplying the Nr. and Dr. by ex]

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE  putting e= t so that ex dx = dt.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE   1 ≡ A (t-1)2 + Bt(t-1)+Ct ....(1) (on resolving into partial fractions)

To find A, putting t = 0 on both sides of (1), we get A = 1

To find C, put t = 1 and we get C = 1. Thus 1 ≡ (t - 1)2 + Bt (t - 1) + t

Comparing the coefficients of t2 on both sides, we get 0 = 1 + B or B = – 1 

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE = log t – log (t – 1) – {1/(t – 1)} + C

= log ex – log (ex – 1) – {1/(ex – 1)} + c  = x – log (ex – 1) – {1/(ex – 1)} + c

Ex.10 Integrate (3x + 1) / {(x – 1)3 (x + 1)}.

Sol.

Putting x – 1 = y so that x = 1 + y, we get  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

arranging the Nr. and the Dr. in ascending powers of y

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Hence the required integral of the given fraction  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

 

Ex.11 Evaluate the integral  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Sol.

Let x = sec2θ  ⇒   dx = 2 sec2θ tanq dθ

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

 

Ex.12 Integrate,  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Sol.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Ex.65 Integrate   Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Sol. 

Put y = tanθ

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

 

Ex.13 Evaluate Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEEdx,  where f(x) is a polynomials of degree 2 in x such that  f(0) = f(1) = 3f(2) = –3. 

Sol.

Let , f(x) = ax2 + bx + c   given, f(0) = f(1) = 3f(2) = –3

∴ f(0) = f(1) = 3f(2) = –3, f(0) = c = –3, f(1) = a + b + c = – 3,  3f(2) = 3 (4a + b + c) = – 3

on solving we get a = 1, b = – 1, c = –3  

∴  f(x) = x2 – x – 3 

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Using partical fractions, we get,  Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

We get, A = –1, B = 2, C = 2

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Ex.14 Integrate 1/(sinx + sin 2x).

Sol.

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Now putting cos x = t, so that – sin x dx = dt, we get

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

The document Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
All you need of JEE at this link: JEE
204 videos|288 docs|139 tests

Up next

FAQs on Integration by Partial Fractions - Mathematics (Maths) Class 12 - JEE

1. What is integration by partial fractions?
Ans. Integration by partial fractions is a technique used to simplify and evaluate integrals of rational functions. It involves breaking down a rational function into simpler fractions, making it easier to integrate.
2. When should I use integration by partial fractions?
Ans. Integration by partial fractions is useful when dealing with integrals of rational functions, where the numerator has a lower degree than the denominator. It allows us to decompose the rational function into simpler terms, making it easier to integrate.
3. How do I decompose a rational function into partial fractions?
Ans. To decompose a rational function into partial fractions, follow these steps: 1. Factorize the denominator completely. 2. Express the rational function as a sum of simpler fractions, with each fraction having a linear factor from the denominator as its denominator. 3. Determine the unknown coefficients by equating the numerator of the original rational function with the sum of the numerators of the partial fractions. 4. Solve the resulting system of equations to find the values of the unknown coefficients.
4. Can integration by partial fractions be used for improper rational functions?
Ans. Yes, integration by partial fractions can also be applied to improper rational functions, where the degree of the numerator is equal to or greater than the degree of the denominator. In such cases, the improper rational function is divided to obtain a proper rational function and a remainder, which can then be integrated separately.
5. Are there any special cases or restrictions when using integration by partial fractions?
Ans. Yes, there are a few special cases and restrictions to keep in mind when using integration by partial fractions: 1. The degree of the numerator must always be less than the degree of the denominator. 2. The denominator should be factorizable into linear and irreducible quadratic factors. 3. If the denominator has repeated linear factors, the partial fractions will include terms with denominators raised to higher powers. 4. If the denominator has repeated irreducible quadratic factors, the partial fractions will include terms with denominators raised to lower powers and linear factors in the numerator.
204 videos|288 docs|139 tests
Download as PDF

Up next

Explore Courses for JEE exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Semester Notes

,

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

,

past year papers

,

Extra Questions

,

ppt

,

mock tests for examination

,

Previous Year Questions with Solutions

,

Important questions

,

Summary

,

MCQs

,

study material

,

Viva Questions

,

Free

,

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

,

Exam

,

video lectures

,

Sample Paper

,

Integration by Partial Fractions | Mathematics (Maths) Class 12 - JEE

,

practice quizzes

,

pdf

,

Objective type Questions

,

shortcuts and tricks

;