√(x-1)/x的不定积分,x√(x+1的不定积分)
2个回答
展开全部
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
∫√(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
追问
第一题好像不对啊
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
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
展开全部
令t=√(x-1) x=t²+1
原式=2∫t²/(t²+1)dt=2∫1-1/(t²+1)dt=2(t-arctant)然后代入t
原式=2∫t²/(t²+1)dt=2∫1-1/(t²+1)dt=2(t-arctant)然后代入t
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询