Question

大神，帮忙解一下高数题啊！！第七题的一三四小问，打勾的那几题

Answer 1

3.1+x^3=(x+1)(x^2-x+1)
用待定系数法：A/(x+1)+(Bx+c)/(x^2-x+1)=1/(x+1)(x^2-x+1)
得A=1/3,B=-1/3,C=2/3
所以∫[1/(1+x^3)]dx =1/3∫(1/(x+1))dx-1/3∫((x-2)/(x^2-x+1))dx
其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c
因为d（x^2-x+1）=(2x-1)dx,所以x-2=1/2(2x-1)-3/2
∫((x-2)/(x^2-x+1))dx=1/2∫(d(x^2-x+1)/(x^2-x+1))-3/2∫(1/(x^2-x+1))dx
其中∫(d(x^2-x+1)/(x^2-x+1))=ln|x^2-x+1|+c
∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根号3/2)^2))
因为∫（dx/(x^2+a^2)）=(1/a)arctan(x/a)
所以∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根号3/2)^2))
=（2/根号3）arctan（(x-1/2)/(根号3/2)）+c
在乘上系数,整理∫[1/(1+x^3)]dx=1/3ln|x+1|-1/6|x^2-x+1|+(1/根号3)arctan((2x-1)/根号3)+c

追答

1.(1+x)/(1-x)^3=A/(1-x)^3+B/(1-x)^2+C/(1-X) [Cx^2+(-B-2C)x+(A+B+C)]/(1-x)^3=(1+x)/(1-x)^3 C=0 -B-2C=1 A+B+C=1 解得 A=2 B=-1 C=0 ∫(1+x)/(1-x)^3dx =∫[2/(1-x)^3-1/(1-x)^2]dx =-2∫(1-x)^(-3)d(1-x)+∫(1-x)^(-2)d(1-x) =(1-x)^(-2)-(1-x)^(-1)+C =1/(1-x)^2-1/(1-x)+C