## jotopia34 2 years ago HELP!!! How do you integrate e^x/1+e^2x*dx

1. slaaibak

can you type it using latex please (the equation tool)?

2. jotopia34

Im so sorry I don't know what latex is?

3. jotopia34

I typed it into Wolfram before, and got the final answer, but I dont know the steps in between

4. slaaibak

Using that equation thing at the bottom. $\int{e^x \over {1 + e^{2x}}} dx$ does it look like that?

5. jotopia34

yes!

6. jotopia34

Oh, I see the equation tool, sorry

7. slaaibak

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

8. jotopia34

thanks soooo much, let me have a minute to digest it now.... lol

9. jotopia34

I don't understand this part: u^2 = e^2x

10. jotopia34

oops, yes I do!

11. slaaibak

haha :) cool

12. jotopia34

Last question, now how to I evaluate it from 0 to 1?....:(

13. slaaibak

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)

14. jotopia34

what is arctan e^1???

15. slaaibak

haha, unfortunately my maths isn't that advanced yet :( but use your calculator to evaluate it.

16. jotopia34

okay, I think I may be able to leave the answer at that since their are no calculators on the quiz

17. slaaibak

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.