∫x/(1-x)dx
原式=-∫x/(x-1)dx
=-∫[(x-1)+1]/(x-1)dx
=-∫[1+1/(x-1)]dx
=-x-∫[1/(x-1)]dx
=-x-ln|x-1|+C
(1+x)/(1-x)的原函数的求解过程
(1+x)/(1-x)=-1-2/(x-1),
∴∫(1+x)dx/(1-x)=-x-2ln|x-1|+c.
求积分∫x/(1+x)^3dx
∫x/(1+x)^3dx =∫(1+x-1)/(1+x)^3dx = ∫1/(1+x)^2d(1+x) - ∫1/(1+x)^2d(1+x) = -1/(1+x) + (1/2)*[1/(1+x)^2] +C
求∫√x-1/x dx
解：
令√(x-1)=t，则x=t²+1
∫[√(x-1)/x]dx
=∫[t/(t²+1)]d(t²+1)
=2∫[t²/(t²+1)]dt
=2∫[(1+t²-1)/(1+t²)]dt
=2∫[1- 1/(1+t²)]dt
=2(t-arctant)+C
=2[√(x-1)-arctan√(x-1)] +C