Question

求不定积分∫[1/(1+x^3)]dx

Answer 1

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