定积分问题? 上限为ln2,下限为0. 求[1-e^(-2x)]^(1/2)dx 的定积分.
1个回答
展开全部
t=[1-e^(-2x)]^(1/2),x=-1/2*ln(1-t^2),
dx=-1/2*1/(1-t^2)*(-2t)dt=t/(1-t^2)*dt
x∈[0,ln2] => t∈[0,√3/2]
∫[1-e^(-2x)]^(1/2)dx
=∫t*t/(1-t^2)dt
=-∫(1-t^2-1)/(1-t^2)dt
=∫1/(1-t^2)dt-∫dt
=1/2∫[1/(1-t)+1/(1+t)]dt-∫dt
=1/2[-ln|1-t|+ln|1+t|]-t+C
=1/2[ln|(1+t)/(1-t)|]-t+C t∈[0,√3/2]
=[ln(2+√3)-√3/2]-0
=[ln(2+√3)-√3/2]
dx=-1/2*1/(1-t^2)*(-2t)dt=t/(1-t^2)*dt
x∈[0,ln2] => t∈[0,√3/2]
∫[1-e^(-2x)]^(1/2)dx
=∫t*t/(1-t^2)dt
=-∫(1-t^2-1)/(1-t^2)dt
=∫1/(1-t^2)dt-∫dt
=1/2∫[1/(1-t)+1/(1+t)]dt-∫dt
=1/2[-ln|1-t|+ln|1+t|]-t+C
=1/2[ln|(1+t)/(1-t)|]-t+C t∈[0,√3/2]
=[ln(2+√3)-√3/2]-0
=[ln(2+√3)-√3/2]
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
