Question

高数有理函数的积分怎么求？

详细过程

Answer 1

∫[(x-3)/(x-1)(x²-1)]dx=∫(x-3)/(x-1)²(x+1)]dx;
将被积函数进行拆项：(x-3)/(x-1)²(x+1)]=[(ax+b)/(x-1)²]+c/(x+1)[下面通分]
=[(ax+b)(x+1)+c(x-1)²]/[(x-1)²(x+1)]=[ax²+(a+b)x+b+cx²-2cx+c]/[(x-1)²(x+1)]
=[(a+c)x²+(a+b-2c)x+(b+c)]/[(x-1)²(x+1)][因为是恒等变换，两边对应项的系数应该相等]
因此有方程组：a+c=0........①；a+b-2c=1...........②；b+c=-3...........③
三式联立求解：①+③得a+b+2c=-3........④；④-②得：4c=-4，∴c=-1；a=1，b=-2；
∴∫[(x-3)/(x-1)(x²-1)]dx=∫[(x-2)/(x-1)²]dx-∫[1/(x+1)]dx
=∫[(x-2)/(x²-2x+1)]dx-∫[1/(x+1)]d(x+1)
=(1/2)∫[(2x-2)/(x²-2x+1)]dx-∫[1/(x-1)²]d(x-1)-∫[1/(x+1)]d(x+1)
=(1/2)∫d(x²-2x+1)/(x²-2x+1)-∫d(x-1)/(x-1)²-∫[1/(x+1)]d(x+1)
=(1/2)ln∣x²-2x+1∣+[1/(x-1)]-ln∣x+1∣+C
=ln∣(x-1)/(x+1)∣+[1/(x-1)]+C;

Answer 2

那个杭绣的要求是可以求解的，具体的你可以考虑一下

Answer 3

高效有理函数的积分怎么求？当然是按照他的定理方式来做，按照勾选和CN的具体标准来做