一道积分题~

∫(0,x)t^2/(1-t^2)dt
=∫(0,x)[-1+1/(1-t^2)]dt
=∫(0,x)[-1+1/2*1/(1-t)+1/2*1/(1+t)]dt
=[-t-1/2ln|1-t|+1/2ln|1+t|]|(0,x)
=-x-1/2ln|1-x|+1/2ln|1+x|
=x+1/2*ln(|1+x|/|1-x|)

I=∫ t^2/(1-t^2) dt (0,x)
let t= sina
dt = cosa da
t=0, a = 0
t=x, a = arcsin x
I = ∫ (sina)^2/(cosa) da (0, arcsinx)
= ∫ tana sina da (0, arcsinx)
= ∫ 1/ (seca)^2 d(seca) a(0, arcsinx)
= [ -1/ (seca) ] a( 0, arcsinx)
= - x + 0
= -x

解：原式=∫(0,x)[1/(1-t²)-1]dt
=-x+∫(0,x)dt/(1-t²)
=-x+1/2∫(0,x)[1/(1+t)+1/(1-t)]dt
=-x+1/2(ln│1+x│-ln│1-x│)
=1/2ln│(1+x)/(1-x)│-x。

原式=-∫(0→x)（1-t^2)dt/(1-t^2)+∫(0→x)dt/（1-t^2)
=-∫(0→x)dt-(1/2)∫(0→x)d(1-t)/(1-t)+(1/2)∫(0→x)d(1+t)/(1+t)
=-x-(1/2)ln|1-x)+(1/2)ln|1+x|
=-x-(1/2)ln(|1+x|/|1-x|).