Question

√（x-1）/x的不定积分，x√（x＋1的不定积分）

Answer 1

x-1=t²搭做中 x=1-t²
∫√(x-1)/xdx=∫t/(1-t²)d(1-t²)=-2∫t²/(1-t²)dt=-2∫(t²-1+1)/胡祥(1-t²)dt
=2∫1-1/(1-t²)dt=2∫dx-2∫1/[(1+t)(1-t)]dt=2t-∫1/(1+t)+1/(1-t)]dt
=2t-ln|(1+t)/(1-t)|+C

∫x√(x+1)dx =∫x√(x+1)+√(x+1) -√(x+1)dx=∫知山√(x+1)³ -√(x+1)d(x+1)
=5/2√(x+1)^5 -3/2√(x+1)³+C

追问

第一题好像不对啊

Answer 2

引用huamin8000的回答：
x-1=t² x=1-t²
∫√(x-1)/xdx=∫t/(1-t²)d(1-t²)=-2∫t²/(1-t²)dt=-2∫(t²-1+1)/(1-t²)dt
=2∫1-1/(1-t²)dt=2∫dx-2∫1/[(1+t)(1-t)]dt=2t-∫1/(1+t)+1/(1-t)]dt
=2t-ln|(1+t)/(1-t)|+C

∫x√(x+1)dx =∫x√(x+1)+√(x+1) -√(x+1)dx=∫√(x+1)³ -√(x+1)d(x+1)
=5/2√(x+1)^5 -3/2√(x+1)³+C

令核搜扰伍t=√（x-1） x=t²+1
原式=2∫t²/(t²+1)dt=2∫1-1/改李历(t²+1)dt=2(t-arctant)然后代入t