几道关于定积分的问题
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/08 02:39:55
几道关于定积分的问题
![](http://img.wesiedu.com/upload/d/7f/d7f1fd7847e50f1ff56e2bdbc0dffe63.jpg)
![](http://img.wesiedu.com/upload/d/7f/d7f1fd7847e50f1ff56e2bdbc0dffe63.jpg)
∫(1-e^-2x)^(1/2)dx
=∫(t^2-1)^(1/2)/t^2dt (t=e^x)
(t^2-1)^(1/2)/t^2=(t^2-1)^(-1/2)-1/t^2(t^2-1)^(1/2)
∫(t^2-1)^(1/2)/t^2dt
=In|t+(t^2-1)^(1/2)|-∫dt/t^2(t^2-1)^(1/2)
=In|t+(t^2-1)^(1/2)|+sgn(u)∫udu/(1-u^2)^(1/2) (u=1/t)
=In|t+(t^2-1)^(1/2)|-(t^2-1)^(1/2)/t
∴原定积分下上限为1,2
原定积分=In(2+3^(1/2))-3^(1/2)/2
=∫(t^2-1)^(1/2)/t^2dt (t=e^x)
(t^2-1)^(1/2)/t^2=(t^2-1)^(-1/2)-1/t^2(t^2-1)^(1/2)
∫(t^2-1)^(1/2)/t^2dt
=In|t+(t^2-1)^(1/2)|-∫dt/t^2(t^2-1)^(1/2)
=In|t+(t^2-1)^(1/2)|+sgn(u)∫udu/(1-u^2)^(1/2) (u=1/t)
=In|t+(t^2-1)^(1/2)|-(t^2-1)^(1/2)/t
∴原定积分下上限为1,2
原定积分=In(2+3^(1/2))-3^(1/2)/2