Question

Improper integral from 0 to infinity ( e^x/e^2x+) dx

Answer 1

could you solve that problem please guys?

Answer 2

\[\int\limits_{0}^{\infty} \frac{e ^{x}}{e ^{2x}+?????}dx\]

Answer 3

sorry
3

Answer 4

\[\int\limits_{}^{}\frac{e^x}{e^{2x}+3} dx\]
\[\text{ Let } e^x=\sqrt{3} \tan(\theta) => e^x dx= \sqrt{3} \sec^2(\theta) d \theta\]
\[\tan(\theta)=\frac{e^x}{\sqrt{3}}\]
Make right triangle. Label that triangle based on what you are given.
\[Opp=e^x, Adj=\sqrt{3} => Hyp=\sqrt{e^{2x}+3}\]
So we have
\[\int\limits_{}^{}\frac{ \sqrt{3} \sec^2(\theta) d \theta}{3 \tan^2(\theta)+3} \]
\[\sqrt{3} \int\limits_{}^{}\frac{\sec^2(\theta)}{3(\tan^2(\theta) +1)} d \theta\]
\[\frac{\sqrt{3}}{3} \int\limits_{}^{}\frac{\sec^2(\theta)}{\sec^2(\theta)} d \theta= \frac{\sqrt{3}}{3} \theta+C\]
\[=\frac{\sqrt{3}}{3} \tan^{-1}(\frac{e^x}{\sqrt{3}})+C\]
So we have
\[\lim_{b \rightarrow \infty} \frac{\sqrt{3}}{3} \tan^{-1}(\frac{e^x}{\sqrt{3}})|_0^b\]

Answer 5

\[\frac{\sqrt{3}}{3}\lim_{b \rightarrow \infty}(\tan^{-1}(\frac{e^b}{\sqrt{3}})- \tan^{-1}(\frac{1}{\sqrt{3}}))\]

Answer 6

\[\frac{\sqrt{3}}{3}(\frac{\pi}{2}-\tan^{-1}(\frac{1}{\sqrt{3}}))\]
\[\frac{\sqrt{3}}{3}(\frac{\pi}{2}-\frac{\pi}{6})\]

Answer 7

\[\frac{\sqrt{3}}{3}(\frac{3 \pi - \pi }{6})=\frac{\sqrt{3}}{3}(\frac{2 \pi}{6})=\frac{\sqrt{3}}{3}(\frac{\pi}{3})=\frac{\sqrt{3} \pi}{9}\]

Answer 8

Thank you so much but I don't understand it

Answer 9

Which part?

Answer 10

I don't know all of it
I don't get it

Answer 11

why don't you use the substitution in the first step

Answer 12

?

Answer 13

i did use substitution in the first step

Answer 14

for this one you don't need to draw the right triangle
i did anyways because i didn't know how the antiderivative was gonna turn out
but my sub came first

Answer 15

Thank You so much Myininaya
You alway help me that is really kind of you

Answer 16

If you want to ask more questions about this problem, don't be afraid to.

Answer 17

To be honest Myininaya
I don't understand it :(

Answer 18

Do you understand the sub that used?

Answer 19

I don't know why you used (tan and sec)

Answer 20

for it

Answer 21

Ok lets look at it for a second
we have...
\[\int\limits_{}^{}\frac{e^x}{e^{2x}+3} dx\]

\[\int\limits_{}^{}\frac{e^x}{(e^{x})^2+3} dx\]
Can you recall the trig identity:
\[\sin^2(x)+\cos^2(x)=1\]
There was also:
\[\tan^2(x)+1=\sec^2(x)\]
(if you divide that first trig identity I wrote by cos^2(x))
So he have
\[(e^x)^2+3 \text{ \in the denominator }\]
We want to figure out a sub that we can make so we can write this as one term
well we just said:

\[(\tan(x))^2+1=(\sec(x))^2\]
Doesn't this look similar to:
\[(e^x)^2+3\] (well at least the left hand side of that equation anyways, right?)
How can we use this identity here:

What if I wrote :
\[\frac{3}{3}(e^x)^2+3=3(\frac{1}{3}(e^x)^2+1)\]

What if we made the sub \[e^x=\sqrt{3} \tan(\theta)\]
so we will be able to use the trig identity that we wrote
\[3(\frac{1}{3}(\sqrt{3} \tan(\theta))^2+1)=3(\frac{1}{3}(3 \tan^2(\theta)+1)=3(\tan^2(\theta)+1)\]
\[=3\sec^2(\theta)\]

Answer 22

since \[\tan^2(p)+1=\sec^2(p)\]
\[x^2 + a \text{ try something with tan( )}\]
since \[\tan^2(p)=\sec^2(p)-1\]
\[x^2 - a \text{ try something with sec( )}\]
since \[\cos^2(p)=1- sin^2(p)\]
\[a - x^2 \text{ try something with sin( )} \]

---
Now these are really the only three trig sub you need
(you can use the others but you only need these 3)

Answer 23

other form of intregration:
\[\int\limits_{0}^{\infty}\frac{e ^{x}}{3+(e ^{x})^{2}}=\int\limits_{0}^{\infty}\frac{e ^{x}}{3\left[ 1+ (\frac{e ^{x}}{\sqrt{3}})^{2} \right]}=\frac{1}{\sqrt{3}}\int\limits_{0}^{\infty}\frac{e ^{x}}{\sqrt{3}\left[ 1+ (\frac{e ^{x}}{\sqrt{3}})^{2} \right]}=\frac{1}{\sqrt{3}}\tan^{-1} (\frac{e ^{x}}{\sqrt{3}})+C\]

Answer 24

So we had that first one I mentioned
\[x^2+a\]
We can even say instead of this that we have
\[ (f(x))^2+a\]
and since we have this we use tan( ) in our sub

Answer 25

Thank You guys some much
I understood right now thank you again