## The Calculus Primer (2011)

### Part XIII. Special Methods of Integration

### Chapter 51. INTEGRATION OF RATIONAL FRACTIONS

**13—8. Rational Fractions.** As used here, this term refers to a fraction in *x* whose numerator and denominator contain *x* to positive integral powers only—in other words, a fraction whose numerator and denominator are polynomials in *x.* For example,

are rational fractions. If the degree in *x* of the numerator is greater than that of the denominator, as in (c) above, the fraction can be changed, by division, to the form below; thus

We shall now show how to integrate fractions of the form (a) and (b); fractions such as (c) are first reduced by division, and the resulting quotient and remainder may then be integrated.

**13—9. Partial Fractions.** To integrate a rational fraction, we first resolve it, by a well-known method of algebra, into partial fractions. For our purposes, we shall consider only three types, limiting the discussion to the simpler situations; other types may be dealt with by analogous methods, but are more cumbersome and occur less frequently in practice.

**13—10. Case I:** *When the Denominator May Be Factored into Real, Linear Factors, None of Which Are Repeated.* The following examples illustrate both the method of resolving the integrand into partial fractions, as well as the actual integration; the latter is, of course, a relatively simple matter.

EXAMPLE 1. Find

*Solution.* By algebra, we may write:

To find the values of *A* and *B,* we clear of fractions, obtaining:

7*x +* 6 *= A*(*x +* 3) *+ B*(*x*) (1)

In equation (1), let *x* = −3; then 7(−3) + 6 = *B*(−3), or *B* = +5. Similarly, let *x* = 0; then 6 = 3*A,* or *A* = 2.

Hence:

Therefore,

= 2 log *x* + 5 log (*x* + 3) + *C*.

EXAMPLE 2. Find

*Solution.* Factoring the denominator:

3*x*^{2} *–* 11*x −* 4 = (3*x* + 1)(*x −* 4).

Resolving into partial fractions:

Clearing of fractions:

2*x −* 34 = *A*(*x −* 4) + *B*(3*x +* 1).

Letting *x* = +4, *B* = −2; when *x* = − , *A* = 8.

EXAMPLE 3. Find

*Solution.* By the method of partial fractions:

or, *6x*^{2} + 12*x* − 30 = *A*(*x* − *d*)(*x* + 2) + *Bx*(*x* + 2) + *Cx*(*x* − 3). Letting *x =* 0, *A* = 5; letting *x* = 3, *B* = 4; letting *x* = −2, *C* = −3.

Hence,

Therefore:

NOTE. The result may be put into a more compact form by writing 5 log *x* + 4 log (*x −* 3) *−* 3 log (*x +* 2) in the form

log *x*^{5} + log (*x −* 3)^{4} *−* log (*x +* 2)^{3},

or

**EXERCISE 13—6**

*Find the following:*

**13—11. Case II:** *When the Factors of the Denominator Are All of the* *First Degree and Some Are Repeated.* This type of rational fraction will now be integrated.

EXAMPLE 1. Find

*Solution.* Let

Then, clearing of fractions:

5*x*^{2} − 3*x* − 2 = *Ax*(*x* − 2) + *B*(*x* − 2) + *Cx*^{2}*.* (1)

Now, letting *x* = 2, we find *C* = 3; letting *x* = 0, *B* = 1; substituting the values of *B* and *C* so found from (1), we get *A* = 2.

NOTE. An alternative method for finding the values of *A*, *B*, and *C* is to use the method of undetermined coefficients. Accordingly, we rewrite (1) as follows:

5*x*^{2} − 3*x* − 2 = *Ax*^{2} − 2*Ax* + *Bx* − 2*B* + *Cx*^{2},

or, 5*x*^{2} − 3*x* − 2 = (*A* + *C*)*x*^{2} + (*B* − 2*A*)*x* − 2*B*. (2)

Equating coefficients of like powers of *x,* we obtain:

*A* + *C* = 5; *B* − 2*A* = −3; −2*B* = −2.

Solving these three simultaneous equations by the methods of elementary algebra yields *A* = 2, *B* = 1, *C* = 3, as before.

EXAMPLE 2. Find

*Solution.* Let

Then

−2*x*^{2} − *x* − 9 = *A*(*x* + 2)(2*x* + 1) + *B*(2*x* + 1) + *C*(*x +* 2)^{2}

= (2*A* + *C*)*x*^{2} *+* (5*A* + 2*B +* 4*C*)*x +*(2*A + B +* 4*C*).

Hence, equating coefficients of like powers of *x:*

or *A* = 1; *B* = 5; *C* = −4.

Therefore,

EXAMPLE 3. Find

*Solution.* Let

Clearing of fractions:

− 6*x*^{2} + 9*x* − 2 = *A*(*x* − l)^{3} + *Bx*(*x* − l)^{2} + *Cx*(*x* − 1) + *Dx,* (1)

or

− 6*x*^{2} *+* 9*x* − 2 = (*A* + *B*)*x*^{3} + (−3*A* − *2B + C*)*x*^{2}

*+* (3*A* + *B* − C + *D*)*x* − A. (2)

From (1), if *x* = 0, *A* = 2; if *x* = 1, *D* = 1; substituting these values of *A* and *B* in (2), we obtain:

−6*x*^{2} + 9*x* − 2 = (2 + *B*)*x*^{3} + (−6 − 2*B* + *C*)*x*^{2}

+ (6 + *B* − *C* + 1)*x* − 2. (3)

Equating coefficients of like powers of *x* in (3):

0 = 2 + *B*, or *B* = −2;

−6 = −6 − 2*B* + *C*, or *C* = 2*B* = −4.

Hence:

**EXERCISE 13—7**

*Find the following:*

**13—12. Case III:** *When the Denominator Contains One or More Factors of the Second Degree, hut None of Them Repeated.* The following examples show how such a fraction may be decomposed into partial fractions; note that to each non-repeated second-degree factor in the denominator of the given fraction there corresponds a partial fraction of the form

where *A, B, p, q, r* = constants.

EXAMPLE 1. Find

*Solution.* Let

Clear of fractions:

13*x* + 36 = *Ax*^{2} *+* 7*A +* 2*Bx*^{2} − 3*Bx* + 2*Cx* − 3*C*

= (*A +* 2*B*)*x*^{2} + (−3*B* + 2*C*)*x +* (7*A* − 3*C*).

Hence,

*A +* 2*B* = 0; −3*B* + 2*C* = 13; 7*A* − 3*C* = 36.

Solving, *A* = 6, *B* = −3, and *C* = 2;

and

Therefore:

**EXERCISE 13—8**

*Find the following integrals:*

**EXERCISE 13—9**

**Review**

*Verify the following integrations without the use of tables:*

**15.** arc sin *x dx* = *x* arc sin *x* + + *C*