Question

Use a contour to evaluate so x²+2x da (2-1)(x3+1)

Answer 1

$ دموع - إ fo رها C the part of the R to : (2-1) 22 +22 da (a -1) (23+1) (22+27) We considen az fcz) az say (7-1) (73+1) whene C is the Contour consisting of the of the semi - Circle CR of padius R together with real anis from R. The function f(7) and (+3+2) i.e., 7-1 and 73 lie, (741) (72-7+1) = 0 7 = -1, 1+ Bu 2 ... The function f(z) has simple poles at 7 = tt 1+ Bi of which z=1 and 1+31 c. lie the closed Contoun inside Hence, Cauchy's residue theorem -1 > Z = 2 z = we obtain by CR (fce) + S Areas on ㅔ 'cp -R R 2) 2 2771! [ Res(t; 1) tres (f; to si How Res (fil) - lim (7-1) f(7) ZL x 7 1 (72+27) lim (Z - A) (7/1) (73+1) (1+2) 를 (141) (2 - 1731) (z? +27) Resff; 19 1+Bi 2 = lim 2 17-1X7+2)(73W). la feli) 2 (7 - 1ārzi xtfity trail -2 +63 ithi (30) (-1-831') (-2+831) 2 - .

31-3 Now, on on CR 1fiz) 1 - 1 1712 +2121 L 171212) (171-1)(1+1441) | 7-11 173+11 R2 (Rus) (R²+1) x (-ta) (+3) 1 TE we )(1+ R3) Applying me - formula obtain If if(a) dz| LRT 1 - ²) (1 7 1 23) CR .TR to as Roso liml f17) dz -0. CR وليا obtain Therefore, proceeding limit on R. R-40 5 867) s ficas + f(a) da Се -R 3 aTi 2 L 2711 1 - 22 [ X +(59) - %) ED fix ) R f(a) da TA3 ot + Steam 22+27 du Hence, T. (2-1) (93+1)