习题4?1 1. 求下列不定积分: (1)?1x2dx; ?2 解 ?1x2dx??xdx?1?2?1x?2?1?C??1x?C. (2)?xxdx; 3 解 ?xxdx??x2dx?321?13x2?1?C?25x2x?C. (3)? 解 ?1xdx; ?121xdx??xdx??112?1x?12?1?C?2x?C. (4)?x23xdx; 7 解 ?x23xdx??xdx?3173?17x3?1?C?310x33x?C. (5)?1x2dxx; ?52 解 ?1x2xdx??xdx??152?1x?52?1?C??32xx?1?C. m (6)?xndx; n 解 ?xdx??xmdx?nmmn1?1nxm?1?C?mn?mm?nxm?C. (7)?5x3dx; 解 ?5x3dx?5?x3dx?x4?C. 45 (8)?(x2?3x?2)dx;
解 ?(x2?3x?2)dx??x2dx?3?xdx?2?dx?x3?x2?2x?C. 3213 (9)?dh2gh(g是常数); 解 ?dh2gh?12g?h?12dh?12g1?2h2?C?2hg?C. (10)?(x?2)2dx;
解 ?(x?2)2dx??(x2?4x?4)dx??x2dx?4?xdx?4?dx?x3?2x2?4x?C. 31 (11)?(x2?1)2dx; 解 ?(x2?1)2dx??(x4?2x2?1)dx??x4dx?2?x2dx??dx?x5?x3?x?C. 5312 (12)?(x?1)(x3?1)dx; 132323 解 ?(x?1)(x?1)dx??(x?x?x?1)dx??xdx??x2dx??x2dx??dx ?x?x2?3321323255x2?x?C. (13)?(1?x)x(1?x)x4dx; 2 解 ? (14)? 解 ? (15)? 解 ?dx??21?2x?xxdx2dx??(x?12131?2x2?x2)dx?2x2?433x2?255x2?C. 3x?3x?1x?13x?3x?1x?1x222422; 2dx??(3x?1x?12)dx?x?arctanx?C3. 1?xx2dx; x?1?11?x221?x2dx??dx??(1?11?x2)dx?x?arctanx?C.
(16)?(2ex?)dx; x 解 ?(2ex?)dx?2?exdx?3?dx?2ex?3ln|x|?C. xx (17)?(31?x31?x22331?21?x21?x22)dx; 解 ?(?)dx?3?11?x2dx?2?11?x2dx?3arctanx?2arcsinx?C. (18)?e(1?xe?x)dx; x 解 ?ex(1?e?xx)dx??(e?xx?12x1)dx?e?2x2?C. (19)?3xexdx; 解 ?3edx??(3e)dx? (20)?2?3?5?23xxxxxxx(3e)xln(3e)?C?3exxln3?1?C. dx; 解 ?2?3?5?23xx2x()2x52x3dx??[2?5()]dx?2x?5?C?2x?()?C23ln2?ln33ln3. (21)?secx(secx?tanx)dx;
解 ?secx(secx?tanx)dx??(sec2x?secxtanx)dx?tanx?secx?C. (22)?cos2dx; 2 解 ?cos2xx2dx??1?cosx2dx?1?(1?cos2x)dx?12(x?sinx)?C.
(23)?dx; 1?cos2x 解 ?dx??dx?tanx?C. 21?cos2x22cosx (24)?dx; cosx?sinx 解 ?dx??dx??(cosx?sinx)dx?sinx?cosx?C. cosx?sinxcosx?sinx (25)? 解 ? (26)?(1? 解 1x21111cos2xcos2xcos2x?sin2xcos2xcoscos2xsinxsin2xxdx; dx?coscos2cos2x22?x?sin222xxsinxdx??(1sin2x?1cos2x)dx??cotx?tanx?C. )xxdx; 3?541??1???x?2x??xdx??(x4?x)dx?477x4?4x?14?C. 2. 一曲线通过点(e2, 3), 且在任一点处的切线的斜率等于该点横坐标的倒数, 求该曲线的方程. 解 设该曲线的方程为y?f(x), 则由题意得 y??f?(x)?11x, 所以 y??dx?ln|x|?C. x2 又因为曲线通过点(e, 3), 所以有?3?2?1 3?f(e 2)?ln|e 2|?C?2?C, C?3?2?1. 于是所求曲线的方程为 y?ln|x|?1. 2 3. 一物体由静止开始运动, 经t秒后的速度是3t(m/s), 问 (1)在3秒后物体离开出发点的距离是多少？ (2)物体走完360m需要多少时间？ 解 设位移函数为s?s(t), 则s??v?3 t2, s??3t2dt?t3?C. 因为当t?0时, s?0, 所以C?0. 因此位移函数为s?t 3.
(1)在3秒后物体离开出发点的距离是s?s(3)?33?27. (2)由t 3?360, 得物体走完360m所需的时间t?3360?7.11s. 4. 证明函数e212x, eshx和echx都是xxexchx?shx的原函数. 证明 exexex?x?x??e2x. x?x?xchx?shxe?ee?ee?2222exchx?shx 因为(1e2x)??e2x, 所以1e2x是的原函数. 因为 (exshx)??exshx?exchx?ex(shx?chx) 所以exshx是ex?e?xex?e?x?e(?)?e2x, 22x exchx?shx的原函数. 因为 (exchx)??exchx?exshx?ex(chx?shx) 所以echx
xex?e?xex?e?x?e(?)?e2x, 22x ex是chx?shx的原函数.
4-1高等数学同济大学第六版本
