Question

Find the partial fraction decomposition for each of the following rational expressions. Then use the partial fraction to give an antiderivative for each function.

(2-6) (3 + 2)2

Answer 1

sot: x (2-6) (x+2) 2 using pavetial fraction method (2-6) (x+2) 2 -07 2242) + (242) X A (x+2)² + B(x-6) (x-6) (x+2)² o l +2) + (x-6) (x+2)2 (x+2) → 62 09 (2722 (x-6) (x+2)² A OL+2)2 + B(x-6) + c(0-6) (x+2) (x-6) (x+2)2 Po z = A (x2+42 + 4) + BX-6B+C (GCP 4X-12) (x-6) (x + 2)² - (x-6) (x+2)2 Ax+ 4x+4) + (BX-6B)+G(X 4X-12) (x-6) (x+2) =

(x+6) (+2) - = (A+C) x2 + (4A+B-40) x + (4A-6B-126) (x-c) (X+2) 2 en compairing both the sides, we get, » A+C = 0 ->0. → 4A+B-4C = 1 -> ☺ → 4A-6B-120 = 0 -> from eq 0. we get Atc=0 A=-C Now, putting A=-6, in eq @ le we get → 4 (-6) + B - 4C = 1 B- 8C = 1 - 4 And, → 46-c) - 6B-12C =0 - 6B - 160 - 0 → 68 +160 = 0 - © Nou guom eq.@ we get B- 8C = 1 & B = 1+ 8c Now, putting B= 1+8C in eq 6(1+80) + 160 = 0 6t 480 + 160 = 0 6 + 646 = 0 &

→ 640 = -6 C = -6 64 C = - 3 3211 i A= -6 And. B = 1+80 = 1+8 [3] * I+ (z] in Replacing each of the coeffivents partial fraction .. , 1/4 3/33 X-6 (0-6) (0+2)² 3/32 ( +je+ + xta,

Now, entegerating wert a weget, Ja-6)(x+2) (x-6) (x+3) a (x+2) de da= lnxto 10(+6) ++/6+2)* dan 10 (4+2) + c [: J4d2 = tnx +c] 10 (2-) + + (tayo (x+2) + c [: [x"dx = pont +67 = 3 In (2-6) (x+2) (-2+1) (at2) + c 32 10 (2-6) -764) - 3 m (2+2)+ c hi kerak.dz . 1(3-6) fexta) (212) 40