Question

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

Answer 1

show full steps if possible o_o

Answer 2

Ready to work together on this?

Answer 3

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

Answer 4

lol i know its pretty rediculous

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

that sounds wrong lol

Answer 20

ok i know :D

Answer 21

@blisk: I chain ruled it.

Answer 22

Sory

Answer 23

d/dx (2x) = 1

Answer 24

i didnt know you could chain rule backwards

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

your right i confused myself sorry!!!

Answer 30

lol dont worry about it

Answer 31

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.

Answer 32

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?

Answer 33

Yes it is.