不定积分∫根号下(x+1⼀x-1)dx，根号包括整个分式

∫((x+1)/(x-1))^0.5dx
上下同乘(x+1)
=∫((x+1)/(x^2-1)^0.5dx
=∫x/(x^2-1)^0.5dx+∫1)/(x^2-1)^0.5dx
=1/2∫1/(x^2-1)^0.5d(x^2-1)+∫1/(x^2-1)^0.5dx
=(x^2-1)^0.5+ln(x+(x^2-1)^0.5)+C

解：
设 x=sect ==>dx=sect*tant*dt ;
∫√[(x+1)/(x-1)]dx = ∫√[(sect+1)/(sect-1)]*sect*tant*dt
= ∫√[(1+cost)/(1-cost)]*sint/cos²t *dt
= ∫√[(1+cost)²/(1-cost)(1+cost)]*sint/cos²t *dt
= ∫[(1+cost)/sint]*(sint/cos²t) *dt
= ∫(sec²t)dt + ∫(cost/(1-sin²t)dt
= tant + 0.5∫[1/(1+sint) + 1/(1-sint)d(sint)
=tant + 0.5 ln[(1+sint)/(1-sint)]+c

x=sect ==> cost = 1/x; sint=√(x²-1)/x; tant=√(x²-1
因此：
原积分式=√(x²-1) + 0.5ln[(1+√(x²-1)/x)/(1-√(x²-1)/x)] + c
=√(x²-1) + 0.5ln[(x+√(x²-1))/(x-√(x²-1)] + c //对数内分子分母同乘以x+√(x²-1)
=√(x²-1) + 0.5ln[(x+√(x²-1)]² + c
=√(x²-1) + ln|(x+√(x²-1)| + c

前部分好积不说了，后一部分设t=tany,y=arctant,dt=(secy)^2 dy,代入积分再代回
4∫(1/(t^2+1)+1/(t∧2+1)^2）dt=4arctant+4S1/(secy)^4 *(secy)^2 dy
=4arctant+4S(cosy)^2dy
=4arctant+S(1+cos2y)d2y
=4arctant+2y+sin2y+c
=4arctan根号（x+1/x-1）+24arctan根号（x+1/x-1)+2根号（x+1/x-1）/（1+（x+1/x-1））+c

4∫(1/(t^2+1)+1/(t^2+1)^2）dt
t=tany,y=arctant,dt=sec^2ydy
∫1/（t^2+1)^2dt
=∫1/sec^4y*sec^2ydy
=∫cos^2ydy
=1/2∫(1+cos2y)dy
=y/2+cos(2y)/4+C=arctant/2+cos(2arctant)+c
4∫(1/(t^2+1)+1/(t^2+1)^2）dt
=4[arctant+arctant/2+cos(2arctant)]+c
将整体根号=t代入解得答案