# Square Root Indefinite Integral

Hello everyone..

## Homework Statement

∫√((1+(e^-x))^2)dx

2. The attempt at a solution
I first tried to do a u sub and then attempt a trig sub however I cant do anything with the e^-x left in the u sub. Does anyone have another way I can integrate this thing??

Thank you for any suggestions/help!

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Ray Vickson
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Hello everyone..

## Homework Statement

∫√((1+e^-x)^2)dx

2. The attempt at a solution
I first tried to do a u sub and then attempt a trig sub however I cant do anything with the e^-x left in the u sub. Does anyone have another way I can integrate this thing??

Thank you for any suggestions/help!

Before doing anything, try computing the square root.

The integral is within an arc length problem if it matters. However when i integrate (1+(e^-x)^2)^1/2 from 0 to 2 i get roughly 2.2214187 and when i integrate 1+e^-x from 0 to 2 i get 2.8647.
Am I allowed to square root 1+(e^-x)^2 and receive 1+e^-x?

I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))

SammyS
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The integral is within an arc length problem if it matters. However when i integrate (1+(e^-x)^2)^1/2 from 0 to 2 i get roughly 2.2214187 and when i integrate 1+e^-x from 0 to 2 i get 2.8647.
Am I allowed to square root 1+(e^-x)^2 and receive 1+e^-x?

I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))
Well that changes everything.

Then, of course, $\displaystyle \ \sqrt{1+\left(e^{-x}\right)^2}\ne 1+e^{-x}\ .\$ You wouldn't expect the two integrals to give the same result.

Ray Vickson
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I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))

Using too many brackets is almost as confusing as not using enough. Do you mean (a): ##\sqrt{1+(e^{-x})^2}## or (b): ##\sqrt{(1 + e^{-x})^2}##? If you mean (a) you can write it as sqrt(1 + (e^{-x})^2); if you mean (b) you can write it as sqrt((1 + e^{-x})^2). You can even write e^-x without the {} bracket, as you have done, but some people would argue with that.

I thought in your OP you meant (b), and that is why I suggested you do the square root first. In (a) you can still simplify it usefully before tackling the integral.

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Well that changes everything.

Then, of course, $\displaystyle \ \sqrt{1+\left(e^{-x}\right)^2}\ne 1+e^{-x}\ .\$ You wouldn't expect the two integrals to give the same result.

your right, so is there a specific technique of integration I should use for this integral?

Using too many brackets is almost as confusing as not using enough. Do you mean (a): ##\sqrt{1+(e^{-x})^2}## or (b): ##\sqrt{(1 + e^{-x})^2}##? If you mean (a) you can write it as sqrt(1 + (e^{-x})^2); if you mean (b) you can write it as sqrt((1 + e^{-x})^2). You can even write e^-x without the {} bracket, as you have done, but some people would argue with that.

I thought in your OP you meant (a), and that is why I suggested you do the square root first. In (b) you can still simplify it usefully before tackling the integral.

A is what the integral i must find looks like. I truly wish it had looked like B though...

Anyone have an idea?

SammyS
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Anyone have an idea?
Looking at the results from WolframAlpha, the indefinite integral is messy, but not impossible.

I'm still attempting to solve it but haven't gotten anywhere yet.

If I substituted u= e^x in any integral and the e^x didn't cross out can I just let it equal u and multiply 1/u into the original integral?? Or is that illegal?

After three substitutions and a partial fraction I have the answer and it matches wolfram alpha. Thanks for the help

So, just to clarify, the integral is

##\int \sqrt{1+e^{-2x}}\;dx##?

Yes that's correct

Alright, well here's what I did:

##\displaystyle\int\sqrt{1+e^{-2x}}\;dx\\
\displaystyle\int\sqrt{1+e^{-2x}}\cdot\dfrac{\sqrt{1+e^{-2x}}}{\sqrt{1+e^{-2x}}}\;dx\\
\displaystyle\int \dfrac{1+e^{-2x}}{\sqrt{1+e^{-2x}}}\;dx\\
\displaystyle\int \dfrac{1+e^{-2x}}{\sqrt{1+e^{-2x}}}\cdot\dfrac{\sqrt{1-e^{-2x}}}{\sqrt{1-e^{-2x}}}\;dx\\
\displaystyle\int \dfrac{\left(1+e^{-2x}\right)\sqrt{1-e^{-2x}}}{1-e^{-2x}}\;dx##

As a substitution, let ##u=e^{-x},\;du=-e^{-x}dx\Rightarrow dx=-\dfrac{du}{u}##.

##\displaystyle-\int \dfrac{\left(1+u^2\right)\sqrt{1-u^2}}{1-u^2}\;\dfrac{du}{u}##

Then, a trig substitution with ##u = \sin t,\;du=\cos t\;dt##.

##\displaystyle-\int \dfrac{\left(1+\sin^2t\right)\sqrt{1-\sin^2t}}{\sin t\left(1-\sin^2t\right)}\cos t\;dt##

Something tells me there may have been a slightly shorter route in the first set of rewriting, though...

Actually, I take it back. After looking over the first few lines, it doesn't look like stopping after one of the earlier steps gives you an easy integral to work with.

Another way you could possibly do this would be a u= e^-x then a trig sub (tan)

Ray Vickson
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Another way you could possibly do this would be a u= e^-x then a trig sub (tan)

For ##\int \sqrt{1+e^{-2x}}\, dx,## substitute ##e^{-2x} = u.##

I would leave the e^-x squared that way the trig sub would work after u=e^-x

member 587159
Just another approach. Consider the substitution x = ln u. So dx = 1/u du

The integral becomes: ##\int \sqrt{1+u^{-2}}/u du##. From here, the integral should be pretty straight-forward with some trig substitutions.

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SammyS
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