How do you evaluate the integral #int e^x/(root5(e^x-1))dx# from -1 to 1?

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Answer 1 · 2016-07-07T21:50:21

#= 5/4( (e - 1)^(4/5) - (1/e - 1)^(4/5) )#

#int_(-1)^(1) dx qquad color{red}{e^x/(root5(e^x-1))}#

spotting the differentiation pattern, namely that :

#d/dx alpha (e^x - 1)^n =color{red}{n* alpha (e^x - 1)^(n-1) e^x}#

and comparing the terms in red, we see that #n-1 = -1/5, n = 4/5#

and #n alpha = 1# so #alpha = 5/4#

we can say that

#int_(-1)^(1) dx qquad e^x/(root5(e^x-1))#

#= 5/4[ (e^x - 1)^(4/5) ]_(-1)^(1)#

#= 5/4( (e - 1)^(4/5) - (1/e - 1)^(4/5) )#

i'm not sure how more more that can be simplified