How can I integrate 1/(4u^2 +3)

In an attempt to integrate 1/(4e^x + 3e^-x) dx, I've used subsitution to get to 1/(4u^2 +3) du, but can't figure out what to do next. I've tried using subsitution again with z=(4u^2 +3), but then everything gets very messy when I attempt to integrate with respect to z instead of u. I'd really appreciate some help on this one.

1 / (4 * e^(x) + 3 * e^(-x)) * dx =>
dx / (4 * e^(x) + 3 * e^(-x)) =>
dx / (4 * e^(x) + 3 / e^(x)) =>
e^(x) * dx / (4 * e^(2x) + 3)

u = e^(x)
du = e^(x) * dx

du / (4u^2 + 3)

So, good job with the substitution. Now, do you remember or know this trig identity?

sec(t)^2 = 1 + tan(t)^2

3 + 4 * u^2 =>
3 * (1 + (4/3) * u^2)

Let u^2 = (3/4) * tan(t)^2
Therefore, u = sqrt(3/4) * tan(t)
and du = sqrt(3/4) * sec(t)^2 * dt

Now we have:

sqrt(3/4) * sec(t)^2 * dt / (3 * (1 + (4/3) * (3/4) * tan(t)^2) =>
sqrt(3/4) * sec(t)^2 * dt / (3 * (1 + tan(t)^2) =>
sqrt(3/4) * sec(t)^2 * dt / (3 * sec(t)^2) =>
sqrt(3/4) * dt / 3

Now we integrate:

sqrt(3/4) * t / 3 + C

u = sqrt(3/4) * tan(t)
u / sqrt(3/4) = tan(t)
u * sqrt(4/3) = tan(t)
2u / sqrt(3) = tan(t)
arctan(2u / sqrt(3)) = t

sqrt(3/4) * arctan(2u / sqrt(3)) / 3 + C =>
sqrt(3) * arctan(2u / sqrt(3)) / (2 * 3) + C =>
(sqrt(3) / 6) * arctan(2 * sqrt(3) * u / 3) + C =>
(sqrt(3) / 6) * arctan((2 * sqrt(3) / 3) * e^(x)) + C

There you go. It's a little complicated looking, but it works.

It wasn't a slip-up

2 / sqrt(3) is 2 * sqrt(3) / 3 when you rationalize the denominator.

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∫du/(4u²+3) x=2u dx=2du then

∫dx/[2(x²+3)]=(1/2√3)arctan(x/√3) +c
=(1/2√3)arctan(2u/√3) +c

That's a trig function of some sort... Can't remember exactly which one, but maybe look up the integrals of sin, cos and tan and it should be one of those :)

You have to look up the general answer. I believe that one has a tan in it.