- a X To any A fraction of the form ; to any factor of the first degree, as x — b, occurring twice in the denominator there correspond two B с partial fractions, and If x — b occurs three times, - 6 D there is an additional fraction, and so on. (x-7) quadratic factor x2+px+q there corresponds a partial fraction Px+Q of the form i if the factor xo+px +q occurs twice, x2 + px +? Rx+S there is a second partial fraction, and so on. (x2+px +9) 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. 5x - 11 Example 1. Separate into partial fractions. 2x2+x-6 Since the denominator 2x2+x-6=(x+2)(2x-3), we assume 5x - 11 A B + 23c2+x-6 3+2 2.-3 where A and B are quantities independent of a whose values have to be determined. Clearing of fractions, 5x—11=A(2x – 3) + B(x+2). Since this equation is identically true, we may equate coefficients of like powers of Q; thus 2A+B=5, -3A+2B=-11; whence A=3, B=-1. 5x – 11 3 1 2x2+x--6 +2 2x - 3 ma+n into partial fractions. A B + .(1). We might now equate coefficients and find the values of A and B, but it is simpler to proceed in the following manner : X-a .(1); 3+3 Since A and B are independent of x, we may give to w any value we please. In (1) put -a=0, or x=a; then A=ma+n; atb mb-n putting a+b=0, or x=-6, B= a+b ma+n 1 ma+n mb-n + (x-a)(x+b) a+b 2+ 23x – 11x2 Example 3. Resolve into partial fractions. A B C + + (2x— 1)(3+2)(3 — ) 2x-1 3 ... 23x – 11x2=A(3+x)(3-2) +B(2x − 1)(3-X) + C(2:— 1)(3+2). By equating the coefficients of like powers of x, or putting in succession 2x-1=0, 3+3=0, 3—3=0, we find that A=1, B=4, C=-1. 23x – 11x2 1 4 1 + 3x2+x-2 Example 4. Resolve into partial fractions. С + + (2-2)2 (1-2x) 1-23 (2-2) :: 3x2+x—2=A(2—2)2+B(1–2x)(x-2) + C(1–2x). Let 1-2x=0, then ; let 3—2=0, then C=-4. 5 3 1 4 42-19 Example 5. Resolve into partial fractions. (2c2+1)(x-4) 42 – 19x Ax+ B с i 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 (2c2+1)(2—4) 22+1 2-4 464. 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. 6x3 +522—7 Example. Resolve into partial fractions. 3x2 – 2x – 1 8x -4 3x2–2x-1 5 1 and + ; 3x2 - 2x - 1 3x+1 -1 6x3 +5X2—7 5 1 2x+3+ + 3x2-25-1 3x+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. 3x2+x-2 Example. Find the general term of when ex (x - 2)2 (1–2x) 1 4 1 4 --1-2) +(1-1)-(-1) Hence the (r+1)th or general term of the expansion is 2 5 1 + 466. The following example sufficiently illustrates the method to be pursued when the denominator contains a quadratic factor. ; 7 + 3 Example. Expand ascending powers of a. (1 + 2) (1 + x2) and find the general term. 7 + 20 A BX + c + 1+30 1+ 22 then A= 3; 3 4 - 3x + = 3(1 + 3)-1 + (4 - 3x)(1+ 22)-1 +(4 – 3x) {1 – 2 + + - ......+(-1)px2p+...}. is 4(-1); therefore in the expansion the coefficient of 2* is 3 +4(-1). (2) If r is odd, the coefficient of an in the second series is – 3(- 1)?, and the required coefficient is 3(-1) ? – 3. r+1 EXAMPLES XLIV. d. Resolve into partial fractions : 1. 7 2 - 1 46 + 13x 2. 22 – 10x + 13 5. (3C — 1) (22 – 5x + 6) 9 6. 7. (x - 1) (3 + 2)2 1+ 3x + 2x2 3. (1 – 2 x)(1 – 22) 2x3 + x2 – 3 – 3 X(3 – 1) (2x + 3) 24 – 3 x3 – 3 22 + 10 ( + 1)(3-3) 8. 26 x2 + 208 x (2 — 114 2 22 – 11x + 5 (x – 3)(x2 + 2 x 5) 5 23 + 6x2 + 5x (22 – 1)(x + 1)3 10. 11. Find the general term of the following expressions when expanded in ascending powers of x. 1+ 3 x 5x + 6 X2 + 7x + 3 12. 13. 14. 1 + 11x + 282 (2 + x) (1 - 3) 22.+ 7 x + 10 2x - 4 4 + 3x + 2 x2 15. 16. (1 - 2) (1 - 2x) (1 – 2) (1 + x 2 2c2) 3 + 2 x – 22 4 + 7 x 17. 18. (1 + x)(1 – 4 x)2 (2 + 3x)(1 + x)2 2 x + 1 19. 20. 1- x + 2x2 (x - 1)(2c2 + 1) (1 – 2 )3 1 3 – 2 x2 21. 22. (1-ax)(1- br)(1-cx) (2 – 3 x + x2) |