令√((x+1)/(x-1))=t
则x=(t^2+1)/(t^2-1)=1+2/(t^2-1)
则dx=-2/(t^2-1)^2*2tdt=-4t/(t^2-1)^2dt
原式=∫(t^2-1)/(t^2+1)*t*(-4t)/(t^2-1)^2dt
=-4∫t^2/[(t^2+1)(t^2-1)]dt
=-2∫[1/(t^2+1)+1/(t^2-1)]dt
=-2arctant-2∫dt/[(t+1)(t-1)]
=-2arctant-∫[1/(t-1)-1/(t+1)]dt
=-2arctant-ln|t-1|+ln|t+1|+C
再代回去
再问： 可是书后给的答案是这个，，，，x>-1时,ln|x+(x^2-1)^(1/2)|-arcsin1/x+C；x-1时，原式=∫1/x*(x+1)/√(x^2-1)dx 令x=1/sint 则原式=∫sint*(1/sint+1)/√(1/sin^2(t)-1)*(-cost/sin^2(t))dt =∫sint(1+sint)/cost*(-cost/sin^2(t))dt =-∫(1+sint)/sintdt =-∫dt/sint-∫dt =∫d(cost)/(1-cos^2(t))-t =1/2ln|(1+cost)/(1-cost)|-t+C =ln|(1+cost)/sint|-t+C =ln|(1+√(1-1/x^2))/(1/x)|-arcsin(1/x)+C =ln|x+√(x^2-1)|-arcsin(1/x)+C 另一种情况类似，只是第一步去根号有点不一样。 不过我的方法应该也没问题，如果没算错的话和答案应该是“一样的”，只是表达式不一样。
