求不定积分∫(dx)/√(1+e^2x)=?
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/13 15:23:18
求不定积分∫(dx)/√(1+e^2x)=?
如题,求详解QAQ!
如题,求详解QAQ!
![求不定积分∫(dx)/√(1+e^2x)=?](/uploads/image/z/15936052-4-2.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%28dx%29%2F%E2%88%9A%281%2Be%5E2x%29%3D%3F)
答:
设t=√ [1+e^(2x) ]
t^2=1+e^(2x)
e^(2x)=t^2-1
2x=ln(t^2-1)
x=(1/2)ln(t^2-1)
原式
=∫ 1/t d [(1/2)ln(t^2-1)]
=(1/2) ∫ (1/t)*[ 1/(t^2-1) ]*2t dt
=∫ 1/(t^2-1) dt
=∫ 1/[(t-1)(t+1)] dt
=(1/2)* ∫ 1/(t-1) -1/(t+1) dt
=(1/2)* [ln(t-1) -ln(t+1) ] +C
=(1/2) ln [(t-1)/(t+1)]+C
=(1/2)*ln { [√(1+e^2x) -1 ] / [ √(1+e^2x) +1 ] } +C
设t=√ [1+e^(2x) ]
t^2=1+e^(2x)
e^(2x)=t^2-1
2x=ln(t^2-1)
x=(1/2)ln(t^2-1)
原式
=∫ 1/t d [(1/2)ln(t^2-1)]
=(1/2) ∫ (1/t)*[ 1/(t^2-1) ]*2t dt
=∫ 1/(t^2-1) dt
=∫ 1/[(t-1)(t+1)] dt
=(1/2)* ∫ 1/(t-1) -1/(t+1) dt
=(1/2)* [ln(t-1) -ln(t+1) ] +C
=(1/2) ln [(t-1)/(t+1)]+C
=(1/2)*ln { [√(1+e^2x) -1 ] / [ √(1+e^2x) +1 ] } +C