Question

高数题，求解！！在线等

第四题

Answer 1

斗罗大陆二百五二十五五二十五

Answer 2

可以用待定系橘判数法

答案档伍如如图所示行启

Answer 3

先化茄薯含拍简：

(x³+1)/[x²(x-1)]
=x³/[x²(x-1)] + 1/[x²(x-1)]
=[x/颤老者(x-1)] + [A/x² + B/x + C/(x-1)]
=[1 + 1/(x-1)] + [A(x-1) + B * x * (x-1) + C * x²]/[x²(x-1)]
=[1 + 1/(x-1)] + [Ax - A + Bx² - Bx + Cx²]/[x²(x-1)]
=[1+1/(x-1)] + [(B+C)x² + (A-B)x - A]/[x²(x-1)]
可见，-A = 1, A-B = 0, B+C=0
所以，A= -1, B = -1, C = 1
那么，化简后的结果为：
= 1 + 1/(x-1) - 1/x² - 1/x + 1/(x-1)
= 1 + 2/(x-1) - 1/x - 1/x²
所以，这个积分等于：
= ∫dx + 2∫dx/(x-1) - ∫dx/x - ∫dx/x²
= x + 2ln(x-1) - ln(x) + 1/x + C