I need some help checking & evaluating this:
$$\huge\int\limits_{0}^{\infty}\frac{e^x}{e^{2x}+3}\ dx=[\frac{1}{\sqrt{3}}\tan^{-1}(\frac{e^x}{\sqrt{3}})]_0^\infty$$

Question

anonymous · Answer

thats right
solved it be letting e^x=u?

anonymous · Answer

u = e^x
du = e^x dx:

anonymous · Answer

Yep, and this is arctangent basically: $\large\frac{1}{u^2+\sqrt{3}}$

anonymous · Answer

But ugh...  the evaluation...

anonymous · Answer

sorry what is ugh?

anonymous · Answer

:-P  It's an expression of disgruntlement

fwizbang · Answer

arctan(infinity)= pi/2.

anonymous · Answer

Grr is more angry frustrated, ugh is more fatigued frustrated

anonymous · Answer

ok :D

anonymous · Answer

But that's e$^{\infty}$

fwizbang · Answer

One infinity is much like another here....

anonymous · Answer

Order of operations @fwizbang , agreed?

anonymous · Answer

Oh so you're saying it doesn't matter, point.

anonymous · Answer

*calculating*

fwizbang · Answer

You only need top worry about "different" infinities when you're dividing things.

anonymous · Answer

$\large\frac{\pi}{3\sqrt3}$?

fwizbang · Answer

why the 3 in the denom?(outside the root?)

anonymous · Answer

Greatest common denominator right?

anonymous · Answer

(27)^(1/2) = 9^(1/2) * (3)^(1/2)

fwizbang · Answer

$$\tan^{-1} (e^x/\sqrt{3}) -> \pi/2$$

the root 3 makes no difference.

anonymous · Answer

Ty for the help gentlemen!  :-)