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HELP!!! How do you integrate e^x/1+e^2x*dx

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can you type it using latex please (the equation tool)?
Im so sorry I don't know what latex is?
I typed it into Wolfram before, and got the final answer, but I dont know the steps in between

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Using that equation thing at the bottom. \[\int{e^x \over {1 + e^{2x}}} dx\] does it look like that?
Oh, I see the equation tool, sorry
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
thanks soooo much, let me have a minute to digest it now.... lol
I don't understand this part: u^2 = e^2x
oops, yes I do!
haha :) cool
Last question, now how to I evaluate it from 0 to 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)
what is arctan e^1???
haha, unfortunately my maths isn't that advanced yet :( but use your calculator to evaluate it.
okay, I think I may be able to leave the answer at that since their are no calculators on the quiz
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.

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