sin(x)dx/(sin^3(x)+cos^3(x))求不定积分

汪红星回答：

　　∫sinxdx/(sinx^3+cosx^3)

　　=∫dx/sinx^2(1+cotx^3)

　　=-∫dcotx/(1+cotx^3)

　　cotx=u

　　=-∫du/(1+u^3)

　　=(-1/6)ln|u^2-u+1|+(1/√3)arctan[(2u-1)/√3]+(1/3)ln|u+1|+C

　　=(-1/6)ln|cotx^2-cotx+1|+(1/√3)arctan[(2cotx-1)/√3]+(1/√3ln|cotx+1|+C

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

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

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

　　=(1/6)∫dx^2/x^2(1-x+x^2)+...

　　=(1/6)∫(x-1)dx^2/[x^2(1-x+x^2)(x-1)]+..

　　=(1/6)∫dx^2/(x-1)(1-x+x^2)-(1/6)∫dx^2/x^2(x-1)+...

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

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

　　=(1/3)ln(x-1)-(1/6)∫d(x^2-x+1)/(x^2-x+1)+(1/6)∫dx/(1-x+x^2)-(1/3)ln(x-1)/x+...

　　=(-1/6)ln|x^2-x+1|+(1/√3)arctan[(2x-1)/√3]+(1/3)ln|x+1|+C