求定积分 其下限为ln2 上限为2ln2 ∫[1/√(e^x -1)]dx 答案是π/6.帮看看求

问题补充：

求定积分,其下限为ln2,上限为2ln2, ∫[1/√(e^x -1)]dx,答案是π/6.帮看看求定积分,其下限为ln2,上限为2ln2,∫[1/√(e^x-1)]dx,答案是π/6.帮看看是我错了还是答案错了

答案：

首先,你在换元后没有转换积分限

其次,x = ln(1 + u^2),dx = 2u/(1 + u^2) du,并不是你写的2/u du

Let u^2 = e^x - 1 then 2u du = e^x dx

when x = ln2,u = sqrt(e^ln2 - 1) = sqrt(2 - 1) = 1

when x = 2ln2,u = sqrt(e^2ln2 - 1) = sqrt(e^ln4 - 1) = sqrt(4 - 1) = sqrt(3)

integrate(ln2->2ln2) 1/sqrt(e^x - 1) dx

= integrate(1->sqrt(3)) 1/u * 2u/e^x du

= integrate(1->sqrt(3)) 2/(u^2 + 1) du

= 2arctan(u) |(1->sqrt(3))

= 2arctan(sqrt(3)) - 2arctan(1)

= 2Pi/3 - 2Pi/4

= Pi/6