Question

这个不定积分如何计算？

Answer 1

简单计算一下即可，答案如图所示

追答

Answer 2

let

(3x+6)/[(x-1)^2.(x^2+x+1)]≡A/(x-1) +B/(x-1)^2+(Cx+D)/(x^2+x+1)

=>

3x+6≡A(x-1)(x^2+x+1) +B(x^2+x+1)+(Cx+D)(x-1)^2

x=1,        =>B=9/3 =3

3x+6≡A(x-1)(x^2+x+1) +B(x^2+x+1)+(Cx+D)(x-1)^2

两边求导

3≡A(x^2+x+1)+A(x-1)(2x+1) +B(2x+1)+2(Cx+D)(x-1)+C(x-1)^2

x=1

3=3A+3B

1=A+B

A+3=1

A=-2

3x+6≡A(x-1)(x^2+x+1) +B(x^2+x+1)+(Cx+D)(x-1)^2

coef. of x^3

A+C=0

-2+C=0

C=2

coef. of constant

-A+B+D =6

2+3+D=6

D=1

(3x+6)/[(x-1)^2.(x^2+x+1)]≡-2/(x-1) +3/(x-1)^2+(2x+1)/(x^2+x+1)

∫(3x+6)/[(x-1)^2.(x^2+x+1)] dx

=∫[-2/(x-1) +3/(x-1)^2+(2x+1)/(x^2+x+1)] dx

=-2ln|x-1| -3/(x-1) +∫(2x+1)/(x^2+x+1)] dx

=-2ln|x-1| -3/(x-1) +∫d(x^2+x+1)/(x^2+x+1) 

=-2ln|x-1| -3/(x-1) +ln|x^2+x+1| +C