How Do You Solve the Integral of (1-e^-x)^(1/2)?

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The integral of (1-e^-x)^(1/2) is discussed, with a proposed substitution of t^2 = 1 - e^(-x) to simplify the problem. One participant provides a complex solution involving logarithmic and square root terms, indicating the difficulty of the problem for an introductory physics class. Another contributor agrees on the complexity and suggests that the problem might be better suited for a higher-level calculus course. The discussion highlights the challenges of solving this integral and the various approaches taken by participants. The integral can indeed be solved, but it requires advanced techniques.
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Homework Statement


what is the integral of (1-e^-x)^(1/2)?
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The Attempt at a Solution

 
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What do you think about the substitution t^2=1-e^(-x)= ?

ehild
 
I think it is
<br /> \frac{\sqrt{1-e^{-x}} \left(e^{x/2} x-2 \sqrt{e^x-1}+2 e^{x/2} \log<br /> \left(-e^{-x/2} \sqrt{e^x-1}-1\right)\right)}{\sqrt{e^x-1}}<br />

Is that what you got as well?

By the way, I find this a rather hard calculus problem for an introductory physics class.
 
CompuChip said:
I think it is
<br /> \frac{\sqrt{1-e^{-x}} \left(e^{x/2} x-2 \sqrt{e^x-1}+2 e^{x/2} \log<br /> \left(-e^{-x/2} \sqrt{e^x-1}-1\right)\right)}{\sqrt{e^x-1}}<br />

Is that what you got as well?

By the way, I find this a rather hard calculus problem for an introductory physics class.

I was too lazy to check either formula. I left e-x unchanged, used the substitution 1-e-x =t2,

dx=\frac{2tdt}{1-t^2}

I also got the logarithm of a fraction plus a constant times the square root. The problem can be solved, and you are right, it should rather be at "Calculus and Beyond"

ehild
 
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