into partial fractions, whose denominators are a + 1, x + 2, if we add together the partial fractions, and compare the several terms of the numerator of the resulting fraction with the corresponding and identical terms of the primitive fraction, we shall get B+ C+ D=6, A- B-3C+2D=0, A+6B-C- D=2, 6A+3C-2D=6; and by solving these equations, we shall find 4=1, B=1, C=2, and D=3; and, therefore, the partial fractions required are fractions two classes rious. 675. Fractions which are rational, and the terms of Algebraical whose numerators and denominators are arranged according distinto the powers of a symbol such as a, have sometimes been guished into distinguished into two classes, according as the highest as proper power of a in their numerators was not less or less than the and spuhighest power of x in their denominators: fractions belonging to the first class have been called spurious, those belonging to the second proper, designations which, though not very appropriate, are frequently convenient, from superseding the necessity of a more specific description : all spurious fractions may be converted, by actual division, into a rational quotient involving no inverse power of e and a remainder which is a proper fraction, inasmuch as the division may be always continued without the introduction of negative powers of a, until the index of the highest power of x in the remainder is less than the index of the highest power of x in the divisor. When, therefore, the primitive fraction is proper, the partial fractions, whose algebraical sum is equal to it, must be proper likewise: and when the primitive fraction is spurious, we commence by reducing it to a quotient involving positive powers of x, and a remainder which is a proper fraction; and this remainder is subsequently resolved into proper partial fractions in the same manner as in the first case. If, therefore, a factor of the denominator of the primitive proper fraction be a+ba, the corresponding partial If the factor be (a+bx)", the corresponding partial which may be transformed into the equivalent form a+a1 (a + bx)+a ̧ (a + bx)2 + ...a,- 1 (a + bx)"−1 (a + bx)" whose numerator may be made to involve the same suc cession of powers of a with the former: the separation of its several terms, striking out the common factors in each, gives us the n partial fractions partial frac 676. The examples which we have given, sufficiently Process for determining shew how very embarrassing the determination of such in- independeterminate coefficients would become if their number was dently the considerable, from the number and complexity of the tion corresequations whose solutions are required: the following con- ponding siderations will enable us to determine them successively, signed facwithout the necessity of having recourse to the solution tor like of such equations. M Let be a proper fraction, of the kind we are consi N dering, and let one of the factors of N be a+bx, or to an as a+bx. it will follow, therefore, that M- AQ is divisible by a+be without a remainder, inasmuch as P is a rational numerator which involves no inverse power of a: if we make, therefore, a + bx=0, or x=a, we shall find P = or in other words, M-AQ=0, when x=a: if we sup Example. pose, therefore, the values of M and Q when a=a, to be represented by m and q respectively, we shall have which is the value of the indeterminate quantity A, which was required to be found: and in a similar manner, the values of the numerators of the partial fractions corresponding to the other simple factors of the denominator of the primitive fraction may be determined. As an example, let it be proposed to resolve the α In this case Ma2, Q1 = (x + 2) (x+3) and a1 = −1: therefore, = + Qe (x + 1) (x + 2) (x+3) x+2 In this case M = x2, Q2 = (x + 1) (~ +3) and a2 = — therefore, -2: In this case, M= x2, Q;= (x + 1) (x+2) and a,=-3: Q3 · 2(x+2) 2(x+3) 677. If two or more factors of the denominator of the Similar primitive fraction be identical with each other, forming for deter a factor of the form (a + bx)", we shall have = process mining the partial fractions corresponding to a factor such as (a+bx)"−1 4 ̧Q+4 ̧Q (a+bx)+... A„-,Q (a+bx)"¬1+P (a+bx)" (a+bx)". and, therefore, Q (a+bx)" P= M−A ̧Q−A ̧Q (a+bx) − ............... . — A„ - 1 (a + bx)" a fraction whose numerator is completely divisible by (a + bx)"; consequently, in the first place, M-AQ is divisible by a + bæ, since every other term of the numerator involves a + ba: therefore, m — 409 m = 0, and A = -· P = M1— A ̧ Q—A ̧ Q (a+bx)..... — A„ -, Q (a+bx)" − 2 |