乘以(-x)加到第二行上,得到 111 ∫(x)=(x-1) 1)(x-2)(x-3)(x-a1234 (x-1)(x-2)(x-3)(x-4) dt= tant+C (2)分部积分 arctan r dr arctan- 1+x21-x2 r=sint arctan r Sin r arctan r dt 1 sin't /1-3 d(cos t r arctan r +C. 6.(1)对非负整数 a?2cdz=/(kT +a)sinda (k丌+x)(1+cos(2x)d 2k+1) 所以 (2k+1)2 sina dr (2)记S0=/ sintppdt,S1=/ tainted,则对非负整数k, (k+1)r (+列如甲Pd=k8+8¦± (−x) \11þ,  f(x) = (x − 1)(x − 2)(x − 3)(x − 4)   1 1 1 1 1 2 3 4 1 22 3 3 4 2 1 23 3 3 4 3   = 12(x − 1)(x − 2)(x − 3)(x − 4). . 5. (1) Z dx (1 − x 2 ) 3/2 x=sin t = Z 1 cos2 t dt = tan t + C = x √ 1 − x 2 + C (2) ©ÜÈ© Z arctan x (1 − x 2 ) 3/2 dx = arctan x x √ 1 − x 2 − Z 1 1 + x 2 x √ 1 − x 2 dx x=sin t = x arctan x √ 1 − x 2 − Z sin t 1 + sin2 t dt = x arctan x √ 1 − x 2 + Z 1 2 − cos2 t d(cost) = x arctan x √ 1 − x 2 + 1 2 √ 2 ln √ 2 + √ 1 − x 2 √ 2 − √ 1 − x 2 + C. 6. (1) éšKê k, Z (k+1)π kπ x sin2 x dx = Z π 0 (kπ + x) sin2 x dx = 1 2 Z π 0 (kπ + x)(1 + cos(2x)) dx = π 2 (2k + 1) 4 , ¤± Z nπ 0 x sin2 x dx = Xn−1 k=0 π 2 (2k + 1) 4 = π 2n 2 4 . (2) P S0 = Z π 0 |sin t| p dt, S1 = Z π 0 t|sin t| p dt, KéšKê k, Z (k+1)π kπ x|sin x| p dx = Z π 0 (kπ + x)|sin x| p dx = kπS0 + S1, 2