Question

Indefinite integral wrt x of 1 over (1+x^2)^2

(Take x = Tan u)

Answer 1

\[\int\limits_{}^{}1/(1+x ^{2})^{2}\]

Answer 2

Shouldn't it be ∫1/(1+x2)2dx ?

Answer 3

1+tanu ^2 = sec^2

tanu=x
sec^2 = dx

Answer 4

\[If x = \tan x, \int\limits\limits 1/(1+\tan^2x)^2dx = \int\limits\limits 1/\sec^2xdx = \int\limits\limits \cos^2 xdx = -\sin^2x +C \]

Did I miss anything?

Answer 5

sec^2/sec^4 = 1/sec^2 = cos^2

Answer 6

the derivative of tanx

Answer 7

-sin^2x+C

Answer 8

cant tell; it ran off the screen

Answer 9

there should be a tan-1(x) invloved in the integral I believe

Answer 10

Here's the whole expression:

If x=tanx,∫1/(1+tan2x)2dx=∫1/sec2xdx=∫cos2xdx=−sin2x+C

Answer 11

That's not a 2*x in the expression. Should be tan^2 x

Answer 12

http://www.wolframalpha.com/input/?i=int%281%2F%281%2Bx^2%29^2%29&a=*C.int-_*IntegralsWord-

Answer 13

Tangent you made a mistake early on and carried it through all your work. You didn't substitute dx=sec^2 u du

Answer 14

Thanks for that link, Amistre. And thank you for pointing that out, Chaguanas. Can you illustrate the correct, step-by-step solution? (This is a fun problem.)

Answer 15

cos^2(u) = 1/2 + cos(2u)/2 ; integrate that

Answer 16

I wonder: Why do you need to substitute not only for x, but also for dx?

Answer 17

when you change variables; you have to calibrate for it

Answer 18

you cant just up and alter equation; yo have to make them equal

Answer 19

Tangent, it is the process called u-substitution. You are thinking about algebraic substitution which is something else.

Answer 20

the dx part carries information that needs to be adjust for

Answer 21

substitutions are like packing up the house and moving down the street; but if you dont modify your dx part, its like leaving the dog behind :)

Answer 22

(This is SO much better than studying alone!!!) I noticed estudier's clue that x should be set = to tan x, and I tried to employ trigonometric substitution. Admittedly I haven't had much practice in these types of integrals, so thanks for helping me along.

Question: If you set x = tan^2 u, dx = sec^2 u du, then doesn't this happen?\[\int\limits 1/(1+\tan^2u)^2*\sec^2udu=\int\limits1/(\sec^2u)^2*\sec^2udu=\int\limits1/\sec^2u*du?\]

P.S. Good analogy, Amistre.

Answer 23

(Gah!)

\[\int\limits1/\sec^2u du\]

This is the last term.

Answer 24

Yes. That is what you integrate.

Answer 25

I think Amistre64 png is mostly right, bit off towards the end.

Answer 26

Using the previous substitution dx=sec^2 u du, du = dx/sec^2u.

∫1/sec2udu = ∫cos^2udu = ∫cos^2u/sec^2u dx = ∫cos^4udx

But maybe you don't need to go that far, integrating with respect to x with a function that involves u?

Answer 27

Where I'm mostly stuck is getting to arctan. That part doesn't make too much sense (maybe because I didn't pay enough attention to integration tables?)

Answer 28

wolfram did it :) I felt lazy and didnt want to retype all of it :)

Answer 29

Tangent just leave it at\[\int\limits_{}^{}\cos ^{2}u\]Convert this with a double angle formula which is easy to integrate.

Answer 30

@ amistre64 I am only going to post integrals that Wolfram can't do from now on:-)

Answer 31

Oh no, you don't! (D:)

In all honesty Wolfram really helped.

Now to brush up on my trig

Answer 32

Hmm..nobody seems too keen on this one
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e049a8f0b8b35337e2a1700