Question

这个定积分怎么求啊

Answer 1

简单计算一下即可，答案如图所示

Answer 2

如下图所示，用凑微积分的方法做，反常积分用极限来做

Answer 3

分享一种“简捷”解法。设e^(1-x)=t。∴e^(1+x)=e²/t，e^(3-x)=e²t，dx=-dt/t。
原式=(1/e²)∫(0,1)dt/(1+t²)=…=π/(4e²)。

Answer 4

∫<1, +∞> dx/[e^(1+x)+e^(3-x)]
= ∫<1, +∞> e^x dx/[e^(1+2x)+e^3]
= (1/e)∫<1, +∞> =-de^x /[e^(2x)+e^2]
= (1/e)(1/e)[arctan(e^x/e)]<1, +∞>
= (1/e^2)(π/2-π/4) = π/(4e^2)

Answer 5

∫<1, +∞> dx/[e^(1+x)+e^(3-x)]
= ∫<1, +∞> e^x dx/[e^(1+2x)+e^3]
= (1/e)∫<1, +∞> =-de^x /[e^(2x)+e^2]
= (1/e)(1/e)[arctan(e^x/e)]<1, +∞>