How do you show whether the improper integral #int e^x/ (e^2x+3)dx# converges or diverges from 0 to infinity?

Question

11286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-25T18:31:41

Assuming that the intended integral is #int e^x/(e^(2x)+3) dx# see below.

Integration by substitution will get an arctan whose argument involves #e^x#. As #xrarroo#, the argument will #rarroo#, so arctan #rarr pi/2#

Let #u = e^3# so #du = e^x dx# and the integral becomes

#int 1/(u^2+3) du#

Now let #u = sqrt3 t# to get

#1/sqrt3 int 1/(t^2+1) dt = 1/sqrt3 tan^-1(t)#

Where #t = u/sqrt3 = e^x/sqrt3#

So #int e^x/(e^(2x)+3) dx = 1/sqrt3 tan^-1 (e^x/sqrt3)#

I have omitted the details to properly express the calculation of an improper integral. I assume that the indefinite integral was the difficulty.