fraction of the form x-a x-b, occurring twice in the denominator there correspond two partial fractions, and If x-b occurs three times, B and so on. To any x-b there is an additional fraction, quadratic factor x2+px+q there corresponds a partial fraction Pr+Q of the form ; if the factor x2+px+q occurs twice, x2+px + q there is a second partial fraction, Rx+S and so on. Here the quantities A, B, C, D,.. P, Q, R, S are all independent of x. We shall make use of these results in the examples that follow. A ; to any factor of the first degree, as whence Example 2. Resolve с (x-b)2 Assume Example 1. Separate into partial fractions. Since the denominator 2x2+x-6=(x+2) (2x-3), we assume B 5–11 A where A and B are quantities independent of x whose values have to be determined. Clearing of fractions, 5 −11 2x2+x-6 = 5x-11=4(2x-3)+B(x+2). Since this equation is identically true, we may equate coefficients of like powers of x; thus 2A+B=5, −3A+2B=−11; A=3, B=-1. 5 −11 3 2x2+x-6x+2 ...... = mx + n (x−α)(x+b) mx+n (x−α)(x+b) = A B + •'. mx+n=A(x+b)+B(x−a) .(1). We might now equate coefficients and find the values of A and B, but it is simpler to proceed in the following manner: 1 2x-3 into partial fractions. Since A and B are independent of x, we may give to x any value we please. In (1) put x-a=0, or x=a; then A= putting x+b=0, or x=—b, B= ... Assume mx+n (x− a)(x+b) Example 3. Resolve = ma+n a+b 1 a+b ( into partial fractions. 23x-11.x2 A B + C + (2x−1)(3+x)(3−x) 2x-1 3+x 3 ... 23x-11x2=A(3+x)(3−x)+B(2x−1)(3−x) = Example 5. Resolve Assume 23x-11x2 (2x-1)(9-x2) ; = = ma+n x-a +C(2x−1)(3+x). By equating the coefficients of like powers of x, or putting in succession 2x-1=0, 3+x=0, 3-x=0, we find that 1 4 A=1, B=4, C=-1. Example 4. Resolve into partial fractions. 3x2+x-2 B с Assume + + (x-2)2(1—2x) 1-2x x-2 (x − 2)2 3 .. 3x2+x-2=A(x-2)2+B(1−2x) (x-2)+ C(1-2x). Let 1-2x=0, then A= 3 let x-2=0, then C=-4. To find B, equate the coefficients of x2; thus 5 3 3-A-2B; whence B= —— +mb-n). 1 3x2+x-2 5 4 3(x-2) (x-2)2 into partial fractions. Ax+B+ 42-19x с ; (x2+1)(x-4) x2+1 x-4 (1); Let x=4, then C=-2; equating coefficients of x2, 0=A+ C, and A=2; equating the absolute terms, 42=−4B+ C, and B=-11, 42-19x 2x-11 2 (x2+1)(x-4) x2+1 X- -4 and 464. In all the preceding examples the numerator has been of lower dimensions than the denominator; if this is not the case, we divide the numerator by the denominator until a remainder is obtained which is of lower dimensions than the denominator. Example. Resolve By division, 6x3+5x2-7 6x3+5x2-7 = = == =2x+3+ 8x-4 =2x+3+ into partial fractions. 8x-4 3x2-2x-1 5 1 + 3x+1 X- 1 465. We shall now explain how resolution into partial fractions may be used to facilitate the expansion of a rational fraction in ascending powers of x. Example. Find the general term of panded in a series of ascending powers of x. By Ex. 4, Art. 463, we have 3x2+x-2 1 5 3x2+x-2 4 (x-2)2 1 5 + =-(1-2)+(-)-(1-1) (2-x)2 when ex Hence the (r+1)th or general term of the expansion is 2r 5 r+ (- 3 + 1 - 1 - - 11 ) 6 2r 2+ " 466. The following example sufficiently illustrates the method to be pursued when the denominator contains a quadratic factor. Example. Expand and find the general term. Assume 1. Let 1 + x = 0, + 1 7+ x (1 + x)(1 + x2) 7+ x A = + (1 + x)(1 + x2) 1 + x 4. 6. · ·. 7 + x = A(1 + x2) + (Bx + C)(1 + x). 7x-1 5x+6x2 in ascending powers of x. Resolve into partial fractions: A+ C, whence C = 4; 2. Bx + C 1 + x2 ...... To find the coefficient of x": (1) If is even, the coefficient of x in the second series g0 is 4(-1); therefore in the expansion the coefficient of x is g" 3 + 4( − 1)2. (2) If r is odd, the coefficient of x in the second series is r-1 r+1 — 3(− 1) 2, and the required coefficient is 3(-1) 2 — 3. · x210x+13 (* − 1)(x2 – 5x+6) 9 (x − 1)(x+2)2 ; EXAMPLES XLIV. d. 46 + 13 x 12x2 - 11x - 15 5. 1... 7. •+(− 1)2x2+.....}. 8. 12. 10. 15. 17. 26 x2 + 208 x (x2 + 1)(x+5) 1+ 3x 19. 3 x3 8 x2+10 (x − 1)4 21. Find the general term of the following expressions when expanded in ascending powers of x. 13. 2 x 4 (1 − x2)(1 − 2 x) 3+2x-x2 (1 + x)(1 − 4 x)2 2x + 1 (x-1)(x2+1) 9. 11. 1 (1 − ax)(1 − bx)(1 − cx) 5x+6 (2 + x)(1 − x) 16. 18. 11x+5 (x-3)(x2 + 2 x − 5) 20. 22. 2 x2 5x3+6x +52 (x2 − 1)(x + 1)3 4+3x+2x2 (1-x)(1 + x2x2) 4+7x 1−x+2x2 (1 - x)8 3- 2x2 |