数学一高等数学习题集大集合

)?c ??earctane??(x?2x22e1?e2?xln(1?x2)?3x?0七. 设f(x)??2 , 求?f(x)dx. ?xx?0?(x?2x?3)e解.РРРРРРРРРРРРРР??(xln(1?x2)?3)dx??f(x)dx??Р2?x(x?2x?3)edx???12?122xln(1?x)?[x?ln(1?x2)]?3x?cx?0? ??2 2x?02?x??(x?4x?1)e?c?1考虑连续性, 所以Р c =－1+ c1, c1 = 1 + c Р?12?122xln(1?x)?[x?ln(1?x2)]?3x?cx?0?f(x)dx??2 2x?02?x??(x?4x?1)e?1?c?xР八. 设f'(e)?asinx?bcosx, (a, b为不同时为零的常数), 求f(x). 解. 令t?e，x?lnt, f'(t)?asin(lnt)?bcos(lnt), 所以Рx 17Р Р f(x)?[asin(lnx)?bcos(lnx)]dx = Р?x[(a?b)sin(lnx)?(b?a)cos(lnx)]?c 2九. 求以下不定积分: 1.Р32x4?xdx ?解. 令x?2sintР323222x4?xdx?32sintcostdt??32(1?cost)costdcost ???3233214 =?cost?cos5t?c?(4?x2)2?(4?x2)2?cР35532.Р53??x2?a2dx(a?0) x解. 令x?asectРx2?a2atantdx(a?0)??asecttant?a?tan2tdt?atant?at?c xasect22 =x?a?aarccosa?c x3.Р?ex(1?ex)1?e2xdxРex1?e2x解.