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Mathematics (Maths) Class 12

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G. Integration Of Rational Functions Using Partial Fractions

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  

Doc: Integration by Partial Fractions JEE Notes | EduRev
If we now reverse the procedure, we see how to integrate the function on the right side of this equation   Doc: Integration by Partial Fractions JEE Notes | EduRev
To see how the method of partial fractions works in general, let's consider a rational function  Doc: Integration by Partial Fractions JEE Notes | EduRev
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) = Doc: Integration by Partial Fractions JEE Notes | EduRev
where an ≠ 0, then the degree of P is n and we write deg (P) = n.
If f is improper, that is, deg(P) Doc: Integration by Partial Fractions JEE Notes | EduRev 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

Doc: Integration by Partial Fractions JEE Notes | EduRev    where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required.


Ex.53 Evaluate Doc: Integration by Partial Fractions JEE Notes | EduRevdx.


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
Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

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.
Doc: Integration by Partial Fractions JEE Notes | EduRev
These constants can be determined as in the following example.


Ex.54 Evaluate  Doc: Integration by Partial Fractions JEE Notes | EduRevdx.


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.

Doc: Integration by Partial Fractions JEE Notes | EduRev
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  Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

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) Doc: Integration by Partial Fractions JEE Notes | EduRev

By way of illustration, we could write  Doc: Integration by Partial Fractions JEE Notes | EduRev


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


Ex.55 Evaluate  Doc: Integration by Partial Fractions JEE Notes | EduRevdx

Sol.

The first step is to divide. The result of long division is  Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev
 

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) Doc: Integration by Partial Fractions JEE Notes | EduRev 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

Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

Ex.56 Evaluate  Doc: Integration by Partial Fractions JEE Notes | EduRevdx


Sol.

Since x3 + 4x = x(x2 + 4) can't be factored further, we write  Doc: Integration by Partial Fractions JEE Notes | EduRev

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   Doc: Integration by Partial Fractions JEE Notes | EduRev

In order to integrate the second term we split it into to parts   Doc: Integration by Partial Fractions JEE Notes | EduRev

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.

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

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
Doc: Integration by Partial Fractions JEE Notes | EduRev
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.57 Write out the form of the partial fraction decomposition of the function

Doc: Integration by Partial Fractions JEE Notes | EduRev
 
Sol.

Doc: Integration by Partial Fractions JEE Notes | EduRev

 

Ex.58 Evaluate  Doc: Integration by Partial Fractions JEE Notes | EduRev

Sol. The form of the partial fraction decomposition is  Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral  Doc: Integration by Partial Fractions JEE Notes | EduRev

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  Doc: Integration by Partial Fractions JEE Notes | EduRev

Ex.59 EvaluateDoc: Integration by Partial Fractions JEE Notes | EduRevdx


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

Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Thus A= 1, B1 = – 3, A2 = 1 and B= 0

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev


Ex.60 Evaluate  Doc: Integration by Partial Fractions JEE Notes | EduRev

Sol.

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Ex.61 Evaluate  Doc: Integration by Partial Fractions JEE Notes | EduRev

Sol.

We have  Doc: Integration by Partial Fractions JEE Notes | EduRev , [multiplying the Nr. and Dr. by ex]

Doc: Integration by Partial Fractions JEE Notes | EduRev  putting e= t so that ex dx = dt.

Doc: Integration by Partial Fractions JEE Notes | EduRev   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 

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev = 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.62 Integrate (3x + 1) / {(x – 1)3 (x + 1)}.

Sol.

Putting x – 1 = y so that x = 1 + y, we get  Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Hence the required integral of the given fraction  Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

 

Ex.63 Evaluate the integral  Doc: Integration by Partial Fractions JEE Notes | EduRev

Sol.

Let x = sec2θ  ⇒   dx = 2 sec2θ tanq dθ

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

 

Ex.64 Integrate,  Doc: Integration by Partial Fractions JEE Notes | EduRev

Sol.

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Ex.65 Integrate   Doc: Integration by Partial Fractions JEE Notes | EduRev

Sol. 

Put y = tanθ

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

 

Ex.66 Evaluate Doc: Integration by Partial Fractions JEE Notes | EduRevdx,  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 

Doc: Integration by Partial Fractions JEE Notes | EduRev

Using partical fractions, we get,  Doc: Integration by Partial Fractions JEE Notes | EduRev

We get, A = –1, B = 2, C = 2

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Ex.67 Integrate 1/(sinx + sin 2x).

Sol.

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

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

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

Doc: Integration by Partial Fractions JEE Notes | EduRev

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