Question

∫√(e^x+1)dx 上限ln2下限0

主要是换元时怎么算
∫√(e^x-1)dx 上限ln2下限0
应该是-1
写错题了。。。。。。擦。。。抱歉

Answer 1

换元整体令√(e^x+1)=t
所以x=ln(t^2-1)
原式=∫tdln(t^2-1)
=∫t*2t/(t^2-1)dt
=∫(2t^2-2+2)/(t^2-1)dt
=∫[2+2/(t^2-1)]dt
=2t|（0，ln2）+∫(1/(t-1)-1/(t+1))dt
=2ln2+ln|(t-1)/(t+1)||(0,ln2)

至于你写错了就更简单了
也是令t=√(e^x-1)，t=ln(t^2+1)
原式=∫tdln(t^2+1)
=∫2t^2/(1+t^2)dt
=∫（2t^2+2-2)/(t^2+1)dt
=∫2dt-2∫1/(t^2+1)dt
=2-2arctant|(0,1)
=2-2arctant1
=2-π/2

Answer 2

a=√(e^x+1)
则x=ln2,a=√3
x=0,a=√2
e^x=a²-1
x=ln(a²-1)
dx=2a/(a²-1) da

原式=∫(√2→√3)2a²da/(a²-1)
=2∫(√2→√3)(a²-1+1)da/(a²-1)
=2∫(√2→√3)[1+1/(a²-1)]da
=2∫(√2→√3)[1+1/2*1/(a-1)-1/2*1/(a+1)]da
=2a+ln|a-1|-ln|a+1|(√2→√3)
=2√3-2√2+ln(2-√3)+ln(3-2√2)