不定积分 ∫ x/1-x² dx？ 根号x分之1不定积分

• ∫x/(1-x)dx
• (1+x)/(1-x)的原函数的求解过程
• 求积分∫x/(1+x)^3dx
• 求∫√x-1/x dx

∫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