Question

求解 不定积分 ！！

过程最好详细点！

Answer 1

6.设(x^2+1)/[(x-1)(x+1)^2]=a/(x-1)+b/(x+1)+c/(x+1)^2,
去分母得x^2+1=[a(x+1)+b(x-1)](x+1)+c(x-1)
=[(a+b)x+a-b](x+1)+cx-c
=(a+b)x^2+(2a+c)x+a-b-c,
比较系数得a+b=1,2a+c=0,a-b-c=1.
解得a=1/2=b,c=-1.
所以原式=(1/2)[ln|x-1|+ln|x+1|]+1/(x+1)+c
=(1/2)ln|x^2-1|+1/(x+1)+c.

Answer 2

原式=1/2*∫[1/(x-1)+(x-1)/(x+1)²]dx
=1/2*ln|x-1|+1/2*∫[1/(x+1)-2/(x+1)²]dx
=1/2*ln|x²-1|+1/(x+1)+C

Answer 3

解如下图所示

Answer 4

用待定系数法进行分式裂项：
令(x²+1)/[(x+1)²(x-1)] = A/(x+1)²+B/(x+1)+C/(x-1)
= [ A(x-1)+B(x+1)(x-1)+C(x+1)² ]/[(x+1)²(x-1)]
= [(B+C)x²+(A+2C)x+(-A-B+C)]/[(x+1)²(x-1)]
B+C=1
A+2C=0
-A-B+C=1
A=-1，B=1/2，C=1/2
即：(x²+1)/[(x+1)²(x-1)] = -1/(x+1)²+(1/2)/(x+1)+(1/2)/(x-1)

∫(x²+1)/[(x+1)²(x-1)]dx
= ∫[-1/(x+1)²+(1/2)/(x+1)+(1/2)/(x-1)]dx
= 1/(x+1)+(1/2)ln|(x+1)(x-1)|+c

Answer 5

方法如下图所示，

请认真查看，

祝学习愉快：