Question

求教智者

Answer 1

令√(1-x)/(1+x)=t
则，(1-x)/(x+1)=t²
==> 1-x=t²x+t²
==> 1-t²=(1+t²)x
==> x=(1-t²)/(1+t²)
则，dx=[-2t(1+t²)-(1-t²)*2t]/(1+t²)²dt=-4t/(1+t²)²dt
所以，原式=∫[(1+t²)/(1-t²)]*t*[-4t/(1+t²)²]dt
=∫[(-4t²)/(1-t²)(1+t²)]dt
=-2∫[1/(1-t²)-1/(1+t²)]dt
=-2∫[1/(1-t²)dt]+2∫[1/(1+t²)]dt
=-∫[1/(1-t)+1/(1+t)]dt+2arctant
=∫[1/(t-1)]dt-∫[1/(1+t)]dt+2arctant
=ln|t-1|-ln|t+1|+2arctant+C
=ln|(t-1)/(t+1)|+2arctant+C
=ln|[√(1-x)/(1+x)-1]/[√(1-x)/(1+x)+1]|+2arctan[√(1-x)/(1+x)]+C