therefore involves n coefficients, some of which however may in certain cases have zero values: and the numerator of the righthand member will be also of (n-1) dimensions and will have n coefficients, involving n undetermined constants N1, N2, ... N; by equating, therefore, the coefficients of the same powers of a on both sides of the equation, there will be n different equations, whereby N1, N2,... N, may be determined; and this of course it is possible to do.

Multiplying both sides of (23) by ƒ (x) we have

f(x) f(x)
+ + Nn
x-α1
x-aq

As the two sides are identical, they are the same for all values of x; let therefore x = a1; and since x-a, is a factor of f (x), all the terms of the right-hand side vanish except the first; and that becomes, and must therefore be evaluated by the method of Chapter V. Vol. I; whence we have, when x = α1,

Thus we have determined in definite forms the values of the n numerators N1, N2, N,, and the decomposition of the rational fraction is complete *.

Now for any integral of the form

tion (11),

dx

we have, by equa

* Further remarks on the theory of this subject will be found in Leçons 5, 6, 7, "Cours d'Algèbre Supérieure," by J. A. Serret, 2nde ed. Paris, 1854.

which form is, when a, is real, as convenient as the result admits of: but if a is impossible, then, to avoid the Logarithms of impossible quantities, we reduce as follows:

a—

Let a+B√1, a-ẞ√−1 be a pair of conjugate roots, and let the coefficients of the partial fractions corresponding to these roots, and found as above, be P+Q√−1, p−q✓−1; so that the two partial fractions are

-

let these be compounded into a single fraction with a quadratic denominator; whence we have

In this example the roots of the denominator are 1, -2,

the coefficients of

are respectively 1, -5, and 5;

The roots of the denominator are 0, -2, -4; and since

The roots of the denominator are 1, +√ −2,

---

since

F(x) f'(x)

1

1

✓=2; and 1

=

the coefficients of

3x2-2x+2'

are respectively,√/= 2, 2 + √ = 2;

12

The roots of the denominator are 0, +1, −√−1,

+ √ −4, ✓4; and since

F(x)

=

f'(x)
1

1

5x+15x2+4; the

24' 24

1

dx

1

x +

1

+

24 X

x + √.

Ex. 6.

The roots of the denominator are 1, −1+√−3, −1−√−3

It will be observed that the coefficient of the second fraction corresponding to a pair of conjugate roots is deduced from that of the first by changing the sign of the impossible part.

21.] Let two or more of the roots of f(x) be equal; then the preceding process of resolution does not admit of being applied directly, because in the right-hand member of (23) there will not be n undetermined constants; and consequently the number of unknown constants is not sufficient to render the two numerators identical; in this case we proceed as follows:

Suppose m roots of f(x) to be equal to a1, and the other roots to be am+1, am+2,... an; so that

then if

m

f(x) = (x—a1) (x-am+1) (x — am+2) · · · (x — αn) ; (31)
F(x)
f(x)

is resolved into a series of fractions of the form (23), the numerator of that which has (x-a1)" in the denominator must be of (m-1) dimensions, and involve m undetermined constants; otherwise the equation cannot be an identity; we must therefore suppose

But, for the purposes of integration, it is more convenient, and it is allowable, to assume the numerator of the first partial fraction in the form

M1+M2 (x−α1) + M3 (x — a1)2 + ... +м, (x −α1)−1;

so that

F(x) f(x)

M m α ̧)TM−1; (33)

=

As to the numerators of these partial fractions, those corresponding to simple factors in the denominators may be determined by the method of the preceding articles; and for the determination of M1, M2, ... Mm› let

(x-am+1)(x-am+2) ... (x — α„) = $ (x) ; φ and let (33), which is the numerator of the fraction which has (x-a)" in the denominator, be symbolized by y (x), so that