# Homework Help: Square Root Indefinite Integral

1. Mar 11, 2013

### B18

Hello everyone..
1. The problem statement, all variables and given/known data
∫√((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!

Last edited: Mar 11, 2013
2. Mar 11, 2013

### Ray Vickson

Before doing anything, try computing the square root.

3. Mar 11, 2013

### B18

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?

4. Mar 11, 2013

### B18

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

5. Mar 11, 2013

### SammyS

Staff Emeritus
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.

6. Mar 11, 2013

### Ray Vickson

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.

Last edited: Mar 11, 2013
7. Mar 11, 2013

### B18

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

8. Mar 11, 2013

### B18

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

9. Mar 11, 2013

### B18

Anyone have an idea?

10. Mar 11, 2013

### SammyS

Staff Emeritus
Looking at the results from WolframAlpha, the indefinite integral is messy, but not impossible.

11. Mar 11, 2013

### B18

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

12. Mar 11, 2013

### B18

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?

13. Mar 11, 2013

### B18

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

14. Mar 11, 2013

### SithsNGiggles

So, just to clarify, the integral is

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

15. Mar 11, 2013

### B18

Yes that's correct

16. Mar 11, 2013

### SithsNGiggles

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...

17. Mar 11, 2013

### SithsNGiggles

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.

18. Mar 12, 2013

### B18

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

19. Mar 12, 2013

### Ray Vickson

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

20. Mar 12, 2013

### B18

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

21. Sep 7, 2016

### Math_QED

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.

Last edited: Sep 7, 2016
22. Sep 7, 2016

### SammyS

Staff Emeritus
You have just responded to a thread that's over 3 years old !