Question

improper integral∫_0^∞▒〖e^x/(e^2x+3) dx〗

Answer 1

First we need to evaluate:
\[\int\limits_{}^{}\frac{e^x}{e^{2x}+3} dx\]
\[\int\limits_{}^{}\frac{e^x}{(e^x)^2+3} dx\]
You need to recall a few trig identities and see which one will be useful here:
\[(\sin(\theta))^2=1-(\cos(\theta))^2\]
\[(\sec(\theta))^2=1+(\tan(\theta))^2\]
\[(\tan(\theta))^2=(\sec(\theta))^2-1\]

Answer 2

both terms in the bottom have + in front of it so we will use the 2nd trig identity I mentioned, right?

Answer 3

\[(\sec(\theta))^2=1+(\tan(\theta))^2 => (a \sec(\theta))^2=a^2+(a \tan(\theta))^2=(a \tan(\theta))^2+a^2\]
So we have:
\[(a \sec(\theta))^2=(a \tan(\theta))^2+a^2\]
So what do you think our substitution would look like?

Answer 4

What is a?

Answer 5

\[\text{ and what is } \tan(\theta) ?\]

Answer 6

lol

Answer 7

We have that we are trying to make the bottom \[(a \tan(\theta))^2+a^2 \]
so we can write it as that one term \[(a \sec(\theta))^2\]
So we are trying to figure out the substitution such that
\[(e^x)^2+3=(a \tan(\theta))^2+a^2\]

Answer 8

So what is the substitution that we need here
\[e^x=?\]

Answer 9

\[e^x=a \tan \theta\]

Answer 10

Gj mertsj

Answer 11

now what is a?

Answer 12

\[u=e^x\]
\[du=e^xdx\]
\[\int \frac{du}{u^2+3}\] almost done

Answer 13

ty. I thought someone should reward you for all that hard work.

Answer 14

And satellite likes to remember extra things

Answer 15

\[a=\sqrt{3}\]

Answer 16

So he is going to show you a "short cut"

Answer 17

But my way shows why his way is true

Answer 18

now look in te back of the book to find integrals of the form
\[\frac{du}{u^2+a^2}\] and you will see that it is
\[\frac{1}{a}\tan^{-1}(\frac{u}{a})\]

Answer 19

I think Satellite is a "get ur dun" guy

Answer 20

because all this nonsense is just showing off. look i can integrate this, look i can integrate that...
the entire content of the majority of calc 2 is on the back two pages of the text

Answer 21

You and I, on the other hand, are ladies of refinement and class.

Answer 22

refinement, class, and reinvent the wheel

Answer 23

That too.