Question

improper integral

Answer 1

\[\int\limits_{-\infty}^{\infty}\frac{ e^x }{ (4+2e^x)^{(5/4)} }\]

Answer 2

write it as limits

Answer 3

http://www.wolframalpha.com/input/?i=integral+negative+infinity+to+infinity+e%5Ex%2F%284%2B2e%5Ex%29%5E%285%2F4%29

Answer 4

what I am not getting is the -2^3/4

Answer 5

i used u sub where u=4+2e^x and du=2e^xdx

Answer 6

so I have \[\frac{ 1 }{ 2 }\int\limits \frac{ 1 }{ u^{5/4} }\]

Answer 7

integrate that\[\frac{ 1 }{ 2 }*-4\frac{ 1 }{ u^{1/4} }\]

Answer 8

I'mma let Solomon finish and see what he has.

Answer 9

But, in short, what they did was factor out a 2 from the bottom, make it \(\Large 2^{1/2} \) then it's just the rule of exponents form there :)

Answer 10

*1/4

Answer 11

lost my latex

Answer 12

ok so would my final one still be correct then of \[\frac{ -2 }{ \sqrt[4]{4+2e^{x}} }\]

Answer 13

what a pain:P

Answer 14

chain rule for the inner argument?

Answer 15

Sorry Solomon, but yes, just not completely simplified :)

Answer 16

oh, it is right, we are dividing by -1/4 and times 1/2 we get -2 on top

Answer 17

just these fractions kill my head

Answer 18

Don't forget your limits :D

Answer 19

ok then what am I doing wrong when I break it up into the limits because Im getting zero\[\lim_{n \rightarrow \infty}\frac{ -2 }{ \sqrt[4]{4+2e^{x}} } from -n \to 0 +\frac{ -2 }{ \sqrt[4]{4+2e^{x}} } from 0 \to -n\]

Answer 20

so when I plug that in everything just canles

Answer 21

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\frac{-2}{\sqrt[4]{4+2e^x}}{\Huge|}^{n}_{0}~~+~~\lim_{b \rightarrow ~-\infty}\frac{-2}{\sqrt[4]{4+2e^x}}{\Huge|}^{0}_{b}}\)

Answer 22

I want to choose more variables versus more pain... like to go slow an in order

Answer 23

did I do this correctly, btw ?

Answer 24

Yea, now just solve that, should give you something \( \sqrt{2} \)

Answer 25

I don't want to do this... lo

Answer 26

L

Answer 27

\[0-\frac{ -2 }{ \sqrt[4]{6} }+(\frac{ -2 }{ \sqrt[4]{6} }-0)\]

Answer 28

wouldn't it just cancel?

Answer 29

Did you do the last limit right?

Answer 30

e^0 is 1 and 2+4 is 6

Answer 31

e^infinity is infinity and infinity*2+4 is just infinity and and 1/infinity is 0

Answer 32

\(\Large e^{- \infty} \) is 0?

Answer 33

yup I got that wrong

Answer 34

so\[\frac{ 2 }{ \sqrt[4]{6} }+(\frac{ -2 }{ \sqrt[4]{6} }+\frac{ 2 }{ \sqrt[4]{4} })\]?

Answer 35

So that leaves you with \(\Large \frac{-2}{(4^.25)} \) which is \(-sqrt{2} \) :)

Answer 36

*
\(\sqrt{2} \)

Answer 37

So yes, those will cancel and leave that rad 2 :P

Answer 38

ok how do I go from \[\frac{ 2 }{ 4^{.25} }\]

Answer 39

multiply top and bottom by 4^.75 right but how does that simplify to 2^.5

Answer 40

If you're looking for exact, I don't know how to get there honestly xD

But if decimal works.. well that equals about the same as \(\sqrt{2}\) when plugged into a calculator

Answer 41

Also, wolfram says so :P
http://prntscr.com/68ok4e

Answer 42

but it wont give you how to get to root 2?

Answer 43

i figured out how to get it to root 2

Answer 44

thanks

Answer 45

You're welcome, good luck :O

Answer 46

*:)