Question

∫(x^2+1)/[(x+1)^2(x-1)]dx

哪位大神可以帮我解解这个题啊？不会啊!

Answer 1

答：

设(x^2+1) / [(x-1)(x+1)^2 ]=(Ax+C) / (x+1)^2 +B/(x-1)
两边同时乘以(x-1)(x+1)^2得：
x^2+1=(Ax+C)(x-1)+B(x+1)^2
=Ax^2+(C-A)x-C+Bx^2+2Bx+B
所以：
A+B=1
C-A+2B=0
B-C=1
解得：A=1/2，B=1/2，C=-1/2
原式
=∫ (1/2)(x-1) /(x+1)^2 +(1/2)/ (x-1) dx
=(1/2) ∫ (x+1-2) /(x+1)^2 +1/(x-1) dx
=(1/2) ∫ 1-2/(x+1)^2 +1/(x-1) dx
=(1/2) x +1/(x+1) +(1/2)*ln |x-1| +C

追问

多谢了！

追答

如有帮助请采纳支持，谢谢，祝你学习进步

Answer 2

这里有些地方你要注意了。
1/[(ax + b)ⁿ(cx + d)] = A/(ax + b)ⁿ + B/(ax + b)ⁿ⁻¹ + ... + X/(ax + b)² + Y/(ax + b) + Z/(cx + d)
(ax + b)/[(cx² + dx + e)(fx + g)] = (Ax + B)/(cx² + dx + e) + C/(fx + g)
如果次方在括号外面的话，就根据第一种情况设
如果括号内是一个多项式的话，就根据第二种情况设，并且分子的次数要比分母少一次
如果括号内是多项式，而括号外又有高次方的话，就两种都用上
即1/[(ax² + bx + c)ⁿ(cx + d)]
= (Ax + B)/(ax² + bx + c)ⁿ + (Cx + D)/(ax² + bx + c)ⁿ⁻¹ + ... + (Xx + Y)/(ax² + bx + c) + Z/(cx + d)

(x² + 1)/[(x + 1)²(x - 1)] = A/(x + 1) + B/(x + 1)² + C/(x - 1)
x² + 1 = A(x + 1)(x - 1) + B(x - 1) + C(x + 1)²
令x = - 1，2 = - 2B ==> B = - 1
令x = 1，2 = 4C ==> C = 1/2
令x = 0，1 = - A - (- 1) + 1/2 ==> A = 1/2
(x² + 1)/[(x + 1)²(x - 1)] = 1/[2(x + 1)] - 1/(x + 1)² + 1/[2(x - 1)]
∫ (x² + 1)/[(x + 1)²(x - 1)] dx = (1/2)∫ dx/(x + 1) - ∫ dx/(x + 1)² + (1/2)∫ dx/(x - 1)
= (1/2)ln|x + 1| + 1/(x + 1) + (1/2)ln|x - 1| + C
= 1/(x + 1) + (1/2)ln|x² - 1| + C
请采纳答案，支持我一下。

追问

(x² + 1)/[(x + 1)²(x - 1)] = A/(x + 1) + B/(x + 1)² + C/(x - 1)

这一步A/(x + 1)是怎么来的？原式只可以拆成两项式啊？