Question

求∫dx/((x^2+1)(x^2+x)不定积分

Answer 1

设：
1/(x^2+1)(x^2+x)=[(ax+b)/(x^2+1)]+(c/x)+[d/(x+1)]
右边通分对应项相等，即可得到：
a=b=d=-1/2,c=1.
此时积分为：
原式
=-（1/2)∫(x+1)dx/(x^2+1)+∫dx/x-(1/2)∫dx/(x+1)
=-(1/2)∫xdx/(x^2+1)-(1/2)∫dx/(1+x^2)-lnx-(1/2)ln(x+1)
=-(1/4)∫d(x^2+1)/(x^2+1)-(1/2)arctanx-lnx-(1/2)ln(x+1)
=-(1/4)ln(1+x^2)-lnx-(1/2)ln(x+1)-(1/2)arctanx+c
=-ln[(1+x^2)^(1/4)*x*(x+1)^(1/2)]-(1/2)arctanx+c.

Answer 2

原式=∫[（ax+b)/(x^2+1)+(cx+d)/(x^2+x)]dx
化简得a=-2/3 b=-1/3 c=2/3 d=1
代入后化简得∫ [ -x/3(x^2+1) - 1/3(x^2+1) +(2x+1)/3(x^2+x) + 2/3(x^2+1) ] dx
(其中∫2/3(x^2+1) dx可化为∫[2/x - 2/(x+1)] dx
再化简得（-1/6)ln(x^2+1) - (1/3)arctan(x) + (1/3)ln(x^2+x) + 2ln(x) - 2ln(x+1)

Answer 3

∫dx/((x^2+1)(x^2+x)dx=
∫[1/x-(1/2)/(x+1)-(x/2)/(x²+1)-(1/2)/(x²+1)]dx
=ln│x│-(1/2)ln│x+1│-(1/4)ln(x²+1)-(1/2)arctanx+C