题目

求x/(x^3+8)的不定积分.

提问时间：2020-10-31

答案

x/(x^3+8)=x/[(x+2)(x^2-2x+4)]=A/(x+2)+(Bx+C)/(x^2-2x+4),
A=-1/6,B=1/6,C=1/3.
x/(x^3+8)=(-1/6)(1/(x+2))+(1/6)(x+2)/(x^2-2x+4).
(-1/6)∫(1/(x+2))dx=(-1/6)(ln|x+2|)
(1/6)∫(x+2)/(x^2-2x+4)dx=
=(1/6)[∫(x-1)/(x^2-2x+4)dx+∫3/[(x-1)^2+3]dx]
=(1/12)ln(x^2-2x+4)+(1/2)∫d(x-1)/[(x-1)^2+3]
=(1/12)ln(x^2-2x+4)+(1/2√3)arctan[(x-1)/√3]
∫x/(x^3+8)dx=
=(-1/6)(ln|x+2|)+(1/12)ln(x^2-2x+4)+(1/2√3)arctan[(x-1)/√3]+C.