1. Improper Integrals

$\int\limits^{\infty}_{0} 4 \frac{e^x}{e^{2x}+3} dx$

Just want to make sure I am starting this correctly.

I rewrote as:

$\int\limits^{\infty}_{0} 4 \frac{e^x}{(e^x)^2+3} dx$

Then did a u-substitution with $e^x$ and got:

$4 \int \frac{u}{u^2+3} du$

Then did another sub with $v= u^2 + 3$ and got:

$2 \int \frac{1}{v} dv$

Integrated and got:

$2 [ln|u^2+3|]$ = $2[ln|e^{2x}+3|]$

Is this how I should be setting this up before I do my limit for my improper integral?

Thanks!

2. $\lim_{t \to \infty} \int_0^{t} \frac{e^x}{(e^x)^2 + 3}\ dx$

Actually: ${\color{red}u = e^x} \ \Rightarrow \ {\color{blue}du = e^x dx}$

(Don't forget to change the limits!)

So: $\lim_{t \to \infty} \int_0^{t} \frac{{\color{blue}e^xdx}}{({\color{red}e^x})^2 + 3} = \lim_{t \to \infty}\int_{1}^{t} \frac{{\color{blue}du}}{{\color{red}u}^2 + 3}$

And this should remind you of the integral: $\int \frac{dx}{x^2 + a^2} = \tfrac{1}{a} \arctan \left(\tfrac{x}{a}\right) + C$

3. Thanks!!

Here is what I got:

$\lim_{t \to \infty} \frac{4}{\sqrt3} [tan^{-1} \frac{e^t}{\sqrt3} - tan^{-1} \frac{1}{\sqrt3}]|^{t}_{0}$

$= \frac{4}{\sqrt3} [\frac{\pi}{2} - \frac{\pi}{6}]$

Look okay?