高数中关于不定积分 的一道题谢谢，，，，，，，，，，，

Answer 1

答：
1/[(x+1)x^3]=a/(x+1)+(bx^2+cx+d)/x^3
=(ax^3+bx^3+cx^2+dx+bx^2+cx+d)/[(x+1)x^3]
所以：
a+b=0
b+c=0
c+d=0
d=1
解得：d=1，c=-1，b=1，a=-1
所以：1/[(x+1)x^3]=-1/(x+1)+(x^2-x+1)/x^3
∫{1/[(x+1)x^3]}dx
=∫[(x^2-x+1)/x^3]dx-∫[1/(x+1)]dx
=∫(1/x-1/x^2+1/x^3)dx-ln|x+1|
=ln|x|+1/x+(-1/2)x^(-2)-ln|x+1|+C
=1/x-1/(2x^2)+ln|x/(x+1)|+C

Answer 2

=ʃ(1/x^3-1/x^2+1/x-1/(x+1))dx=-1/2*x^(-2)+1/x+lnx-ln(x+1)+c