Question

5. Now let's use everything we've learned to solve some more complicated integrals. (a) 23 +2 +63 +1 x"(+1) (b) 229 + 4c - 3. - 3 dar (x + 1)(x+2)

Answer 1

B x + + 22 ze 2 Coe+1) Solcetion : Using partial fractions +x^4G x + 1 A Су +7 ge? (84+1) x²+1 → 823 +82 +6X + 1 A (x (2+1)) + B (X?+1) + Cr2+1) 82 collect up the like terms x3+x2 + 6x+1 ze 3 C Atc)+? (B+D) + x A+B comparing coefficients get A+C B+D = 1 A=6 , B=1; C = -5,D=0 A = 6 - we = 1 B=1 Hence can be written as given integral 5 *8*7*+62+2 de 6 da + + - s(- - 5 g2+1 22 22 (8?+1) ) -5 strane dx + 6 of 12 dx + 5 + 22 dox s ~ Now consider. Som te dog = +5 279 = Llog (x2+1) = s 12 de s 1 22 de Loga s x² da and s = - X So becomes x3+x²+68+1 dae - 5 x?442+1) I Log (x2+1) +6 Loga- to

(b) 2x3+4x38²_3 (x+1)(x+2) da solution : Using partial fractions = + - 243 + 48-34²-3 A 2+B Ca+D (42+11 CQP+2) (0+1) (x2+2) 2083_322+48-3 (AX+B)(x+2) + Cex+D) (ge?+1) = 2x3 –342+42-3 23(A+C) +02 ( B+ D) 78 (2 A+09+ (2B+D) comparing coefficients A+C = 2 B+D = -3 A= 2 2 A+C = 4 B=0 2B+D=-3 С =0 D =-3

22 4+1 2+2) 2x3 - 36 + 4% -3 (2+1) (2+2) so given integral 243 +49-342-3 (42+10 (22+2) can be written as da da 2x 3 su (Q2+13 (&?+2) } - da 3 28 x²+1 s S x 42 de consider 22 x211 da = Log (4+1) s time and da pet = us del = true da » 2. du = da → Sore dx = SE 2u²+2 du du = u²+1 } tant cu) S x² 42 dox = I tanto th be written as can da 293 +48 - 39?-3 (n+1)(x+2) Log (2+1) tante