What is the Derivative of Arctan?

[Saladsamurai]

I know this must be similar...

$$\int \frac{e^x}{1+e^{2x}}$$

should $$u=1+e^{2x}$$?

Casey

 

[bob1182006]

actually this one's a bit trickier.

if you try 1+e^(2x) you won't really go anywhere since the derivative of e^(2x) is 2e^(2x) >< which doesn't appear on the top of the fraction.

So is there any other substitution you can try? one that when derived will give you the quantity that is on top of the fraciton? o.o

 

[Saladsamurai]

I see something, bit I can't figure out how to word it...I know $$e^{2x}=e^{(x)2}$$...or something...

 

[bob1182006]

yes o.o so what other substitution can you try? 1+e^(2x) doesn't work so..

ps.
(this time you won't be able to do the entire quantity on the bottom of the fraction).

 

[Saladsamurai]

actually this one's a bit trickier.

if you try 1+e^(2x) you won't really go anywhere since the derivative of e^(2x) is 2e^(2x) >< which doesn't appear on the top of the fraction.

So is there any other substitution you can try? one that when derived will give you the quantity that is on top of the fraciton? o.o

$$e^x$$ is the only thing that keeps popping into my head...but..

 

[bob1182006]

but? o.o why don't you try it?

when doing integrals it's not weird to try one method for like 1/2 a page then say "this is going nowhere" and going some other way >< from the start...

 

[rocomath]

is this calc 2? b/c all it is the integral of arctan after you u subst. for e^x.

 

[Saladsamurai]

I am...and I am getting...hey wait is this a true statement? $$(e^x)^2=e^{2x}$$

Casey

 

[rocomath]

yes.

$$u=e^{x}$$

$$du=e^xdx$$

what is the integral of arctan?

 

[Saladsamurai]

Ah! It's late...I had myself convinced that $$e^{2x}$$ and $$(e^x)^2$$ were not equal...silly algebra.

Thanks again and goodnight,
Casey

 

[HallsofIvy]

yes.

$$u=e^{x}$$

$$du=e^xdx$$

what is the integral of arctan?

I think you mean "what is the derivative of arctan!

 

[rocomath]

I think you mean "what is the derivative of arctan!

argh! yes actually that would be correct, lol.