Question

integrate (-6e^2t)/(1+e^4t)
iam having lots of trouble with this problem, any formula i try is wrong, any help that includes some steps would be amazing!!!

Answer 1

i wonder if long division would help :)

Answer 2

\[\frac{-6e^{2t}}{1+e^{4t} }\]


-6e^2t + 6e^6t - 6e^10t
-------------------
1+e^4t ) -6e^2t
(-6e^2t - 6e^6t)
--------------
6e^6t
(6e^6t+6e^10)
-------------
-6e^10t
(-6e^10t -6e^14t)
-------------------
.....

Answer 3

\[\frac{-6e^{2t}}{1+e^{4t} }=\sum_{n=0}^{inf}(-1)^{n+1}6e^{(2+4n)t}\]
right?

Answer 4

if the answer is the last part of ur first box, they answer is incorrect. I also typed it into wolfram alpha and it still didnt take it that answer

Answer 5

wolf says another way to write it is -3 sech(2t)

if you can integrate that?

Answer 6

http://www.wolframalpha.com/input/?i=sum+%28%28-1%29%5E%28n%2B1%29e%5E%28%282%2B4n%29t%29%29+from+0+to+infinity

i tool out my 6 and it says this is equal to -1/2 sech (2t) lol

Answer 7

\[6\int \frac{-e^{2t}}{1+e^{4t} }=\int 6\sum_{n=0}^{inf}(-1)^{n+1}e^{(2+4n)t}\]

Answer 8

\[\int 6(-e^{2t} +e^{6t} - e^{10t}+e^{14t}...)dt=6(-\frac{1}{2}e^{2t} +\frac{1}{6}e^{6t} -\frac{1}{10} e^{10t}+\frac{1}{14}e^{14t}...)\]

Answer 9

\[\text{ Let } u=e^{2t} => du=2 e^{2t} dt => -3 du=-6 e^{2t} dt \]
\[\int\limits_{}^{}\frac{-3 du}{1+u^2}=\]
this should help

Answer 10

thats not how orsene would do it :)

Answer 11

\[-3\tan^{-1}(u)+C=-3\tan^{-1}(e^{2t})+C\]

Answer 12

i tried sub but i didnt know how to find it int, once i took the derivative
how e^4t become u^2?

Answer 13

\[u=e^{2t} => u^2=(e^{2t})^2=e^{2 \cdot 2 \cdot t}=e^{4t}\]

Answer 14

By the law of exponents of course :)

Answer 15

But you can try amistre's way or whoever that orsene guy is which is probably amistre anyways...

Answer 16

http://www.wolframalpha.com/input/?i=sum+%28%28-1%29%5E%28n%2B1%29*6e%5E%28%282%2B4n%29t%29%2F%28%282%2B4n%29t%29%29+from+0+to+infinity

i got that too lol

-3 tan-1(2t)
---------- tho i think +C
t

Answer 17

:)

Answer 18

i put a spurious little t in the denominator by accident

Answer 19

little amistre do you mean e^(2t) inside your glamours tan inverse function?

Answer 20

http://www.wolframalpha.com/input/?i=sum+%28%28-1%29%5E%28n%2B1%29*6e%5E%28%282%2B4n%29t%29%2F%28%282%2B4n%29%29%29+from+0+to+infinity

thats better

Answer 21

yes yes yes ... these fat stubby fingers just werent meant for typing

Answer 22

glamorous*

Answer 23

id give your teacher or whoever is giving you this my answer and see f they accept it lol

Answer 24

Are you suppose to use these substitutions:

u = e^t and t= ln u so dt= du/u

Answer 25

i love u guyss thank u!

Answer 26

i used u=e^{2t}

Answer 27

seems to work just peachy! :)

Answer 28

i used .... infinite series :)

Answer 29

I think you get something that gets you an arctangent

Answer 30

amistre was trying to invent a new kind of space travel with some weird math

Answer 31

The pied piper, eh?

Answer 32

zarkon wouldve appreciated it; but jamesj would have shook his head in a contemptible manner and scolded me .... im sure

Answer 33

amistre I approve your message. :)