5.求∫中 解设c=+品+，红∈，则 1 a(x+1)2+bc(x+1)+cc (+1)2 x(x+12 a2+2x+1)+be2+)+ce x(x+1)2 =(a+0r'+(atb+o)r+a x(x+1)2 →a=1;a+b=0→b=-a=-1: 2a+b+c=0→c=-2a-b=-2+1=-1 所以 +那++ 1 从而 ∫-∫-/+女-∫ 1 1 =-+)-/++) 1 =n-n++z++d sin 5 sin 2rdr=cos(5-2r)-cos(5x+2)d 1 1 2x3 cos 3rd(3r)-2x cos7rd(Tz) 1 =后n3r-4m7z+e 数学分析四试题第4页（共8页）5. ¦ R 1 x(x+1)2 dx )  1 x(x+1)2 = a x + b x+1 + c (x+1)2 , ∀x ∈ R, K 1 x(x + 1)2 = a(x + 1)2 + bx(x + 1) + cx x(x + 1)2 = a(x 2 + 2x + 1) + b(x 2 + x) + cx x(x + 1)2 = (a + b)x 2 + (2a + b + c)x + a x(x + 1)2 =⇒ a = 1; a + b = 0 ⇒ b = −a = −1; 2a + b + c = 0 ⇒ c = −2a − b = −2 + 1 = −1. ¤± 1 x(x + 1)2 = 1 x + −1 x + 1 + −1 (x + 1)2 . l Z 1 x(x + 1)2 dx = Z 1 x dx − Z 1 x + 1 dx − Z 1 (x + 1)2 dx = ln |x| − Z 1 x + 1 d(x + 1) − Z 1 (x + 1)2 d(x + 1) = ln |x| − ln |x + 1| + 1 x + 1 + d. 6. ¦ R sin 5x sin 2xdx ) Z sin 5x sin 2xdx = Z 1 2 [cos(5x − 2x) − cos(5x + 2x)]dx = 1 2 Z cos 3xdx − 1 2 Z cos 7xdx = 1 2 × 3 Z cos 3xd(3x) − 1 2 × 7 Z cos 7xd(7x) = 1 6 sin 3x − 1 14 sin 7x + c. êÆ©Û(II)ÁK 1 4 £
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