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II. To every factor of the form (x - a)" corresponds a series of partial fractions of the form

M. M &c. + M,-1.

(x - a): + (x – a)»–1 + Any one of the coefficients as M, is given by the equation

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III. To every factor of the form av + ax + b corre

. M c + N . sponds a fraction at awto. To

5. To determine the constants M and N, the expression (2x + a) – (Mx

= 0

is reduced by successive substitutions of - (a x + b) for yo to the form

Ax + B = 0, and from the conditions A = 0, B = 0, M and N are found.

IV. To every factor of the form (x2 + ax + b)" corresponds a series of fractions of the form

Mv+ N M@+ N, M,-1X + N,-1 (x2 + ax + b)" * (x2 + ax + b)n-1 * &c. x2 + ax + b

To determine M and N let V = Q (2x2 + ax + b)"; then if by the successive substitutions of – (ax + b) for x the equation

U (+ N) Q= 0 be reduced to the form

AX + B = 0, the equations A = 0, B = 0 are conditions for finding M and N. If now we put

U - (Mx + N) Q
na + ax + b

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where U, is necessarily an integral function, we can, from the equation

l; - (11x + N) Q = 0, determine 11 and N, as before, and so in succession for all the other partial fractions.

The fraction having been thus, by one or other of these methods, decomposed into a sum of simpler fractions, each of them may be integrated separately by known processes, and so the whole integral is found.

If the partial fraction be of the form _-.

-aWe have

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(5) Let

V m3 - m - x + 1° Here the denominator contains two equal factors (x - 1)?, and the partial fractions arising from these equal roots are

300-13.--1

and the fraction corresponding to the other factor (0 + 1) is

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(5) Let = 2 (0-1); n being ever

(8)

V

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Here there are in the denominator two equal quadratic factors · (x + 1); the fractions arising from them are

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