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slaaibak Group TitleBest ResponseYou've already chosen the best response.1
can you type it using latex please (the equation tool)?
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
Im so sorry I don't know what latex is?
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
I typed it into Wolfram before, and got the final answer, but I dont know the steps in between
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
Using that equation thing at the bottom. \[\int{e^x \over {1 + e^{2x}}} dx\] does it look like that?
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
Oh, I see the equation tool, sorry
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
e^x = u du = e^x dx u^2 = e^2x \[\int\limits {1 \over 1 + u^2} du\] now letting u = tan k du = sec^2 k dk so it becomes: \[\int\limits {\sec^2 k \over 1 + \tan^2k} dk\] and that equals (because sec^2 k = tan^2 k + 1) \[\int\limits {\sec^2 k \over \sec^2 k} dk = k + C\] but u = tan k k = arctan u so it's arctan u + C but u = e^x so it's arctan e^x + C you could have directly made the substitution e^x = tan u, but this makes one understand it better
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
thanks soooo much, let me have a minute to digest it now.... lol
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
I don't understand this part: u^2 = e^2x
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
oops, yes I do!
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
haha :) cool
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
Last question, now how to I evaluate it from 0 to 1?....:(
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
Since you know the anti derivative, let's call it F, is F(x) = arctan e^x + C, you simply need to evaluate F(1)  F(0) but because you are subtracting the two functions, you don't have to work with the C, because C  C = 0 So it's arctan(e^1)  arctan(e^0) = arctan(e)  arctan(1)
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
what is arctan e^1???
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
haha, unfortunately my maths isn't that advanced yet :( but use your calculator to evaluate it.
 one year ago

jotopia34 Group TitleBest ResponseYou've already chosen the best response.0
okay, I think I may be able to leave the answer at that since their are no calculators on the quiz
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
oh, yes. Just leave it in this format: \[\arctan (e)  \arctan (1)\] Do you understand all the steps? I'll elaborate if there is anything that's poorly explained.
 one year ago
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