不定积分x^4/（1+x^3）dx详细过程

Question

不定积分 x^4/（1+x^3）dx详细过程

Answer 1

∫x^4/（1+x^3）dx= ∫x-x/（1+x^3）dx
用待定系数法：A/(x+1)+(Bx+c)/(x^2-x+1)=x/(x+1)(x^2-x+1)
解得A=-1/3 B=1/3 C=1/3
∫x-x/（1+x^3）dx=∫x+1/3*[1/(x+1)-(x+1)/(x^2-x+1)]dx
其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c
d（x^2-x+1）=(2x-1)dx
所以x+1=1/2(2x-1)+3/2
∫((x+1)/(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
乘上系数,整理
∫[1/(1+x^3)]dx=1/2*x^2+1/3ln|x+1|-1/6|x^2-x+1|-(1/根号3)arctan((2x-1)/根号3)+c