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# Doc: Integration by Partial Fractions JEE Notes | EduRev

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

The document Doc: Integration by Partial Fractions JEE Notes | EduRev is a part of the JEE Course 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

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

where S and R are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required.

Ex.53 Evaluate dx.

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

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

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.

These constants can be determined as in the following example.

Ex.54 Evaluate  dx.

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.

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

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)

By way of illustration, we could write

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

Ex.55 Evaluate  dx

Sol.

The first step is to divide. The result of long division is

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

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

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)  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

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

Ex.56 Evaluate  dx

Sol.

Since x3 + 4x = x(x2 + 4) can't be factored further, we write

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

In order to integrate the second term we split it into to parts

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.

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

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

Sol.

Ex.58 Evaluate

Sol. The form of the partial fraction decomposition is

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

We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral

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

Ex.59 Evaluatedx

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

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

Thus A= 1, B1 = – 3, A2 = 1 and B= 0

Ex.60 Evaluate

Sol.

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

Ex.61 Evaluate

Sol.

We have   , [multiplying the Nr. and Dr. by ex]

putting e= t so that ex dx = dt.

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

= 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

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

Hence the required integral of the given fraction

Ex.63 Evaluate the integral

Sol.

Let x = sec2θ  ⇒   dx = 2 sec2θ tanq dθ

Ex.64 Integrate,

Sol.

Ex.65 Integrate

Sol.

Put y = tanθ

Ex.66 Evaluate dx,  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

Using partical fractions, we get,

We get, A = –1, B = 2, C = 2

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

Sol.

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

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## Mathematics (Maths) Class 12

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