Integrating: Help with \int {\frac {1}{\sqrt {{e^{x}}+1}}} dx

[negatifzeo]

Homework Statement

$\int {\frac {1}{\sqrt {{e^{x}}+1}}} dx$

The Attempt at a Solution

I tried integration by parts, using ${\frac {1}{\sqrt {{e^{x}}+1}}}$ as u and dx as dv. This seemed to make the problem more complicated. Using e^x+1 as a u substitution won't work either. Any help getting on the right path would be greatly appreciated!

 

[rock.freak667]

You could try hyperbolic trig sub. such as e^x=sinhu

 

[negatifzeo]

I don't know what hyperbolic trig sub. is yet. We've just gotten to integration by parts.

 

[rock.freak667]

I don't know what hyperbolic trig sub. is yet. We've just gotten to integration by parts.

Have you done normal trig substitutions? As e^x=tan²u looks like it would work out fine

 

[negatifzeo]

Nope. Maybe it's a problem I'm just supposed to try and not succeed at.

 

[rock.freak667]

Nope. Maybe it's a problem I'm just supposed to try and not succeed at.

Did you learn partial fractions yet? If you did try putting $u^2=e^x+1$ and post back what you tried.

 

[negatifzeo]

I'm not sure of the procedure when using that technique. if u^2=e^x+1, then does the integral become 1/u? Then the answer would just be ln(u)... I don't think that's right

 

[rock.freak667]

I'm not sure of the procedure when using that technique. if u^2=e^x+1, then does the integral become 1/u? Then the answer would just be ln(u)... I don't think that's right

u^2=e^x +1
then 2u du =e^x dx and e^x=?
and so dx =?

 

[negatifzeo]

This problem is driving me nuts. There was a typo earlier, it's actually e^x-1.

Expanding on what you said, if u^2=e^x-1
2udu=e^xdx and dx=(2udu/e^x). Moving all of this stuff around is just fine and dandy but I don't know where to put it back in.