题目

几道求不定积分的题,
1,∫ 1/( (x-2)^2*(x-3) ) dx
2,∫ sinx * sin2x * sin3x dx
3,∫(x^2 - 5x + 9) / ( x^2 - 5x +6 ) dx
4,∫ cosx / ( sinx( 1 + sinx)^2 ) dx
1,1/(x-2) + In| (x-3)/(x-2)| + c
2,1/8*(1/3 * cos(6x) - 1/2 * cos(4x) - cos(2x)) +c
3,x+3In|(x-3)/(x-2)| +c
4,In|sinx/(1+sinx)| + 1/(1+sinx) +c

提问时间：2021-03-26

答案

1、设1/[(x-2)²(x-3)] = A/(x-2)²+B/(x-2)+C/(x-3)解得：A=-1,B=-1,C=1∫dx/[(x-2)²(x-3)]=-∫dx/(x-2)²-∫dx/(x-2)+∫dx/(x-3),有理积分法=1/(x-2)-ln|x-2|+ln|x-3| + C=1/(x-2)+ln|(x-3)/(x-2...