# Nasty integral - which piece of technology to trust?

1. Sep 3, 2010

### tatiana_eggs

Nasty integral -- which piece of technology to trust?

1. The problem statement, all variables and given/known data

indefinite integral: e^x / (e^2x + 9) dx
Technology is allowed to solve it.

2. Relevant equations

3. The attempt at a solution

My trusty TI-89 says (pi/540)arctan((e^3x)/3)

while two different online integral calculators came up with (1/3log(e))*arctan((e^3x)/3)

What do you think? Which is right/should I trust? Should I just do it by hand? If so, what method do you recommend?

Thanks

2. Sep 3, 2010

### ╔(σ_σ)╝

Re: Nasty integral -- which piece of technology to trust?

u=e^x

3. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

Hmm, I don't know if it's just by foggy memory of Calc II but that substitution doesn't seem to help it at all.

4. Sep 3, 2010

### Dickfore

Re: Nasty integral -- which piece of technology to trust?

Make the subst [itex]e^{x} = 3 u[/tex] and your integral will be converted to a table one. none of the answers is correct.

5. Sep 3, 2010

Re: Nasty integral -- which piece of technology to trust?

What relationship is there between $$e^{2x}$$ and $$e^x$$? That should help with the integral and the substitution.

6. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

Thanks you two, I am starting to get it, but I am running into trouble.

So far, 3u=e^x, du=?dx -- should I take du/dx of e^x/3 and get du = e^x dx? But if I do that then I get an e^x in my integral below.

integral: 3u / (9u^2 + 9)

7. Sep 3, 2010

### Dickfore

Re: Nasty integral -- which piece of technology to trust?

what is du?

8. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

The derivative of u. I thought I had to take the derivative of my subsitution and solve for dx to sub that back into my integral.

9. Sep 3, 2010

### Dickfore

Re: Nasty integral -- which piece of technology to trust?

Write what you get for du.

10. Sep 3, 2010

### Hurkyl

Staff Emeritus
Re: Nasty integral -- which piece of technology to trust?

I think your trusty TI-89 calculated exactly what you told it to calculate, rather than what you meant to calculate. Is it in degree mode, perchance, instead of radian mode?

11. Sep 3, 2010

### Dickfore

Re: Nasty integral -- which piece of technology to trust?

yes :)

12. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

so I got du = (e^x)/3 dx. What do I do with this expression? I thought I was supposed to solve for dx, and plug that into my integral with my u's. Is that not right?

integral: 3u / (9u^2 + 9) * 3 / (e^x) du

13. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

Hurkyl, it was in radian mode.

14. Sep 3, 2010

### Dickfore

Re: Nasty integral -- which piece of technology to trust?

Oh, so then the default output of the arctangent function for TI-89 is in degrees.

15. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

Oh wow, I just did it in degree mode and it produced the answer consistent with the back of the book.

16. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

Thanks Statdad, Dickfore and Hurkyl, so very much!

17. Sep 3, 2010

### tatiana_eggs

Re: Nasty integral -- which piece of technology to trust?

Oh and I finally figured out the substitution thing. I realized I didn't need to sub 3u into the numerator

18. Sep 4, 2010

### ╔(σ_σ)╝

Re: Nasty integral -- which piece of technology to trust?

u=e^x
du=e^xdx