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

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Any one of the coefficients as M, is given by the equation

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every factor of the form a2 + ax + b corre

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

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To determine

M and N let V = Q(x2+ ax + b)"; then

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the equations A = 0, B = 0 are conditions for finding M and

N. If now we put

U − (Mx + N) Q

= U1,

x2 + a x + h

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

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determine M 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.

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dx (Mx + N)

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2 (r− 1) {(x − a)2 + B2 }r-1

da

+ (Ma + N) √ {(x − a)2 + ß3}' *

The expression for the last integral will be found in the following Chapter on formulæ of reduction.

(1) Let

U
V

2x+3

x2 + x2 − 2 x

In this case the factors of V are x, x 1, x + 2, and as

d V

=

3x2 + 2x 2,

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Here the denominator contains two equal factors (x − 1)2, and the partial fractions arising from these equal roots are

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

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to 0.

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2x2 + 7x2 + 6x + 2

x2 + 3x2 + 2x2

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The roots of the denominator are 2, 1 and two equal

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U

V

=

x2 dx

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x3 + 5 x2 + 8 x + 4

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+ log (x + 1).

(1) a being even;

U

(8) Let

V

a" (x

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1

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1

+ &c.

-x)'

72

n

N

...

2 (n - 1)

1.2 (n − 1)

...

log (1)

Murphy, Camb. Transactions, Vol. vi.

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The factors of V are + 1, 1, and x2 + 2.

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x2+3x2+2

U

=

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(11) Let-711

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+ x2 + +

Hence

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3x2 + x
(x − 1)3 (x2 + 1)

=

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x.

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(x2 - x+3) (x + 1)18

20

1

3x

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Here there are in the denominator two equal quadratic factors (x2+1); the fractions arising from them are

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