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

Answer 1

Elijah Abram

#= 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

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Answer 2

Adam Adkins

To evaluate the integral (\int_{-1}^{1} \frac{e^x}{\sqrt{5(e^x - 1)}} dx), you can make the substitution (u = e^x - 1), which implies (du = e^x dx). The integral becomes (\int_{0}^{e-1} \frac{1}{\sqrt{5u}} du). This is a standard integral. You integrate (\frac{1}{\sqrt{5u}}) with respect to (u), then substitute back (e^x - 1) for (u), and evaluate the expression at the upper and lower limits of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.