- anonymous

integral(1/((2x)^2+1)dx

- schrodinger

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- anonymous

show full steps if possible o_o

- anonymous

Ready to work together on this?

- anonymous

This one's going to be a bit tricky, I'm afraid.

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## More answers

- anonymous

lol i know its pretty rediculous

- anonymous

Okie dokie. Let's do this. I'll give you a hint: The thing is in the form of a trig identity.

- anonymous

Can you tell me the trig identities for cos^-1 tan^-1 and sin^-1 first? After that, I think you'll know exactly what the answer is :D

- anonymous

don't you have to find the antiderivative of the function?

- anonymous

well i know its arctan its just the steps involved inbetween to find the answer

- anonymous

im just not sure if its u sub or something thats why i wana see it with full steps

- anonymous

@galactic yes you have to find the antiderivative saying integral is another way of saying anti derivative

- anonymous

You know your "X" is 2x. Because of that, you must chain rule it. When you do the derivative of 2x, you get 2. Pull that two out, and because it's on the bottom, it turns to 1/2. So 1/2 tan^-1 (2x) + c should be the answer. But double check with other people to be safe.

- anonymous

If my wording was weird, I meant that normally it'd be just tan^-1 if x^2 was there. But we have (2x)^2 instead of just (x)^2.

- anonymous

shouldnt the 1/2 go on the top and turn into 2arctan(2x)?

- anonymous

@uber i wanted to ask how comes we are using the trig identities. is it because it is one over the function??

- anonymous

@Blisk: Well I mean it's originally just (1/(2(tan^-1(2x)). So you pull the 1/2 to the left.

- anonymous

@galactic; Look at the derivative of tan^-1.

- anonymous

how did you get 1/(2tan^-1(2x))?

- anonymous

Can Someone Help Me After They finish doing this guy!! :D

- anonymous

that sounds wrong lol

- anonymous

ok i know :D

- anonymous

@blisk: I chain ruled it.

- anonymous

Sory

- anonymous

d/dx (2x) = 1

- anonymous

i didnt know you could chain rule backwards

- anonymous

Oh shoot! You're right! I think I made up a rule then. :(

- anonymous

isnt the derivative of tan^-1......sec^2(x)tan^-1?!?

- anonymous

the deriv of tan^-1 is 1/x^2+1

- anonymous

the deriv of tan = sec^2(x) and the deriv of sec is sectan

- anonymous

your right i confused myself sorry!!!

- anonymous

lol dont worry about it

- anonymous

Draw a right triangle having legs of lengths 1 and 2x. Place the unit length segment om the positive x-axis with one end at the origin O. Erect the perpendicular to the x-axis at A(1, 0) and place the other leg on that perpendicular with its lower end at
(1, 0). The other end of that vertical leg is at B(1, 2x). Lrt t be angle AOB. Then
tan(t) = 2x/1 = 2x, and (sec(t))^2 dt = 2 dx =>dx = (sec(t))^2/2 dt.
If we substitute into the integral, we get(1/2)(sec(t))^2 dt/(1 + (tan(t))^2. and we end up integrating dt/2, so that simplifies to t/2. Getting back to original variable x, we have
(1/2) Arctan(2x) + C for the final answer.

- anonymous

would this work too?
S = integral
S1/((2x)^2+1) dx u = 2x
du = 2 dx => 1/2 du = dx
S(1/u^2+1)(1/2) du
1/2S(1/u^2+1) du
1/2 tan^-1(u) Plug U back in
1/2 tan^-1(2x)
is that also correct abtre?

- anonymous

Yes it is.

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