anonymous
  • anonymous
integral(1/((2x)^2+1)dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
show full steps if possible o_o
anonymous
  • anonymous
Ready to work together on this?
anonymous
  • anonymous
This one's going to be a bit tricky, I'm afraid.

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anonymous
  • anonymous
lol i know its pretty rediculous
anonymous
  • anonymous
Okie dokie. Let's do this. I'll give you a hint: The thing is in the form of a trig identity.
anonymous
  • 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
  • anonymous
don't you have to find the antiderivative of the function?
anonymous
  • anonymous
well i know its arctan its just the steps involved inbetween to find the answer
anonymous
  • anonymous
im just not sure if its u sub or something thats why i wana see it with full steps
anonymous
  • anonymous
@galactic yes you have to find the antiderivative saying integral is another way of saying anti derivative
anonymous
  • 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
  • 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
  • anonymous
shouldnt the 1/2 go on the top and turn into 2arctan(2x)?
anonymous
  • anonymous
@uber i wanted to ask how comes we are using the trig identities. is it because it is one over the function??
anonymous
  • anonymous
@Blisk: Well I mean it's originally just (1/(2(tan^-1(2x)). So you pull the 1/2 to the left.
anonymous
  • anonymous
@galactic; Look at the derivative of tan^-1.
anonymous
  • anonymous
how did you get 1/(2tan^-1(2x))?
anonymous
  • anonymous
Can Someone Help Me After They finish doing this guy!! :D
anonymous
  • anonymous
that sounds wrong lol
anonymous
  • anonymous
ok i know :D
anonymous
  • anonymous
@blisk: I chain ruled it.
anonymous
  • anonymous
Sory
anonymous
  • anonymous
d/dx (2x) = 1
anonymous
  • anonymous
i didnt know you could chain rule backwards
anonymous
  • anonymous
Oh shoot! You're right! I think I made up a rule then. :(
anonymous
  • anonymous
isnt the derivative of tan^-1......sec^2(x)tan^-1?!?
anonymous
  • anonymous
the deriv of tan^-1 is 1/x^2+1
anonymous
  • anonymous
the deriv of tan = sec^2(x) and the deriv of sec is sectan
anonymous
  • anonymous
your right i confused myself sorry!!!
anonymous
  • anonymous
lol dont worry about it
anonymous
  • 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
  • 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
  • anonymous
Yes it is.

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