Question

@solomonzelman

Answer 1

How do I find the antiderivative of \(\largey=\Large\frac{1}{1+x^2}\)

@SolomonZelman @freckles @mathmale

Answer 2

ignore the messed up coding

Answer 3

That's a very common integrand. Do you have a table of integrals? If so, y ou should recognize yours among those in the table, very quickly.

Answer 4

um...i don't think so

Answer 5

http://tutorial.math.lamar.edu/pdf/Common_Derivatives_Integrals_Reduced.pdf

Answer 6

tan^-1 x

Answer 7

https://www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&es_th=1&ie=UTF-8&q=table%20of%20basic%20integrals&oq=table%20of%20basic%20integrals&aqs=chrome..69i57j0l4.6360j0j7

I intentionally shared this set of general results in the hope you'd look thru it for the integral in question.

Answer 8

is it tan^-1 x?

Answer 9

Right. I searched for "table of basic integrals." What did nincompoop look up?

While I recognize the right answer in your proposal (StudyGurl), I'd much rather that you write or type or draw it as \[\int\limits_{}^{}\frac{ 1 }{ 1+x^2 }=\tan ^{-1}x+C\]

Answer 10

Ah, ok.

Answer 11

is there any way I can solve it tho without using a table?

Answer 12

In the long run it will save you time and save the time of others if you'd please either draw your results or use Equation Editor (below).

Answer 13

Without a table? The integral you've just used is basic and probably needs to be memorized. You could prove it by finding the derivative of \[\tan ^{-1}x\]

Answer 14

Also, the integral can be proven via trigonometric substitution.

Answer 15

Ideally, you'd know and understand both approaches.

Answer 16

how do I do trignometric substitution to solve it?

Answer 17

Haven't done that for a while, but I'll try to remember the steps involved.

Answer 18

If you are to integrate \[\frac{ 1 }{1+x^2 }\]

Answer 19

Are you with me so far? If not, ask for clarification.

Answer 20

k

Answer 21

\[\frac{ 1 }{ 1+\tan^2u }\]

Answer 22

i'm wth you

Answer 23

Great.
There is a trig identity for \[1+\tan^2x\]

Answer 24

do you know it?

Answer 25

StudyGurl: twice now I have seen that you are working on some other problem, having to do with circumcenter. I am giving you time and effort and do expect that you stick around until we are done.

Answer 26

Take care. Perhaps we could continue later.

Answer 27

I'm sorry. I come back when I get a notification, but I haven't gotten any notifications for this thread...I'm very sorry for you're irritation

Answer 28

your*

Answer 29

and the trig identiy is \(\large \sec^2(x)\)