#1

I need to solve the indefinite integral for: ( x^2 ) / ( 9 + x^6 ) dx

I know the first step is using 'u substitution where u=x^3 and du=3x^2 so that the x^2 on top will cancel out, but I'm lost from there.

Also, I believe the answer is: ( 1 / 9 ) * arctan( x^3/3) + C where arctan is the same as tan inverse.

Can somebody provide a step by step solution while explaining each step?

I know the first step is using 'u substitution where u=x^3 and du=3x^2 so that the x^2 on top will cancel out, but I'm lost from there.

Also, I believe the answer is: ( 1 / 9 ) * arctan( x^3/3) + C where arctan is the same as tan inverse.

Can somebody provide a step by step solution while explaining each step?

#2

The answer is 726X83 ^93i9. It needs no explanation I'm sure.

#3

I never understood why people bring stuff like this to The Pit. in the odd case you get a real answer, it's probably wrong.

#4

The answer is 6. ALWAYS 6.

#5

I need to solve the indefinite integral for: ( x^2 ) / ( 9 + x^6 ) dx

I know the first step is using 'u substitution where u=x^3 and du=3x^2 so that the x^2 on top will cancel out, but I'm lost from there.

Also, I believe the answer is: ( 1 / 9 ) * arctan( x^3/3) + C where arctan is the same as tan inverse.

Can somebody provide a step by step solution while explaining each step?

Step one, close your open internet browser.

#6

I need to solve the indefinite integral for: ( x^2 ) / ( 9 + x^6 ) dx

I know the first step is using 'u substitution where u=x^3 and du=3x^2 so that the x^2 on top will cancel out, but I'm lost from there.

Also, I believe the answer is: ( 1 / 9 ) * arctan( x^3/3) + C where arctan is the same as tan inverse.

Can somebody provide a step by step solution while explaining each step?

( x^2 ) / ( 9 + x^6 ) dx? Simple.

Step 1: Realize you're never going to need to use this, ever, for the rest of your life after highschool/college.

Step 2: Get a hobby.

#7

Math thread. Nao.

And stop trying to act smart.

And stop trying to act smart.

#8

whoa. if you had asked me a year ago, i could have told you the answer easily. google a little bit and you can find some help with it. it's pretty simple once you get going.

....but yeah, most people will never directly use this information in real life.

....but yeah, most people will never directly use this information in real life.

#9

lmao i wasnt trying to act smart. Its a legitament problem and I cant solve it and i need to know how to do it for a calc quiz...

#10

I had my mathematical analisis exam yesterday . I'm trying to solve the bitch

#11

lmao i wasnt trying to act smart. Its a legitament problem and I cant solve it and i need to know how to do it for a calc quiz...

Then find the notes you made in class about integration.

However, if you failed to make notes, you deserve not to know.

#12

i dont know. that sure sounds like a tricky math problem. you know something else im unsure about is how come every single person who answered your thread so far tried to tell you what to do instead of answering your question. maybe theyre ignorant.

#13

I know the first step is using 'u substitution where u=x^3 and du=3x^2 so that the x^2 on top will cancel out, but I'm lost from there.

Also, I believe the answer is: ( 1 / 9 ) * arctan( x^3/3) + C where arctan is the same as tan inverse.

Can somebody provide a step by step solution while explaining each step?

You're close. Notice that you can re-write the bottom as (3^2 +(x^3)^2). When you set u=x^3, you have (1/3) times the integral of: du/(3^2 +u^2). Evaluated, your answer should be something like (1/3)arctan(x^3/3) + C.

Sorry if thats unclear.

#14

Also, I believe the answer is: ( 1 / 9 ) * arctan( x^3/3) + C where arctan is the same as tan inverse.

WRONG! the answer is 6, you've been told.

#15

Go To A ****ing Maths Forum

#16

Then find the notes you made in class about integration.

However, if you failed to make notes, you deserve not to know.

Oh damn. You just got your ass handed to you by Dr. Phil.

#17

I'm learning this stuff now, too.

In FRENCH.

Not that it makes a difference when solving the problems. =P

In FRENCH.

Not that it makes a difference when solving the problems. =P

#18

You're close. Notice that you can re-write the bottom as (3^2 +(x^3)^2). When you set u=x^3, you have (1/3) times the integral of: du/(3^2 +u^2). Evaluated, your answer should be something like (1/3)arctan(x^3/3) + C.

Sorry if thats unclear.

I think this is it . I failed Maths anyway

#19

You're close. Notice that you can re-write the bottom as (3^2 +(x^3)^2). When you set u=x^3, you have (1/3) times the integral of: du/(3^2 +u^2). Evaluated, your answer should be something like (1/3)arctan(x^3/3) + C.

Sorry if thats unclear.

I understand that the integral of: 1/(1 + x^2) is arctan... but I cant figure out how it relates to the (3+x^6) in the denominator.. How did u get the arctan(x^3/3)?

And I have been taking notes. This is a homework problem, so theres no reason why I would have this problem done from class time. Not to mention in class they usually teach you the basics and then homework accelerates the difficulty.

#20

i could show you, but im too lazy and have my own work

#21

I understand that the integral of: 1/(1 + x^2) is arctan... but I cant figure out how it relates to the (3+x^6) in the denominator.. How did u get the arctan(x^3/3)?

And I have been taking notes. This is a homework problem, so theres no reason why I would have this problem done from class time. Not to mention in class they usually teach you the basics and then homework accelerates the difficulty.

1/(1+x^2) is arctan in its simplest form. Let's say that in that case, 1=a and x=u. It could be re-written 1/(a^2+u^2) The full answer is (1/a)arctan(u/a). This works because du=dx. In this case, the integral is a little more complicated. Here, set a=3 and u=x^3. Next, du=3x^2 dx. This eliminates the x^2 while putting a (1/3) on the outside of the integral. Now, rewritten. (1/3) Integral of: du/(a^2 +u^2). Using the previous formula and remembering that a=3 and u=x^3, your answer is (1/3)arctan(x^3/3) + C

#22

I know the first step is using 'u substitution where u=x^3 and du=3x^2 so that the x^2 on top will cancel out, but I'm lost from there.

Also, I believe the answer is: ( 1 / 9 ) * arctan( x^3/3) + C where arctan is the same as tan inverse.

Can somebody provide a step by step solution while explaining each step?

this would have made sense to me 2 years ago when i was doing AS maths. now that im doing economics a uni, i have no idea.

i can differentiate though! its actually used in economics.

sorry to be of absolutely no help to you though

#23

The Pit is not your math teacher.

And you should be thanking GOD that we aren't...

And you should be thanking GOD that we aren't...

#24

First, factor out 9 from the denominator, then move the 1/9 out before the integral. Now, you have x^2/(1+(x^6)/9) dx. set u=(x^3)/3 , since in the original form, u is squared. du= x^2 dx, so you are left with integral of du/(1+u^2). Substitute u (x^3/3) back into the original equation, and add the 1/9 you factored out. du/(1+u^2) is the form for tan inverse/arctan, so you are left with (1/9) tan-1(x^3/3) + C.

I'm in Calc II, and I just learned this not even 2 hours ago.

I'm in Calc II, and I just learned this not even 2 hours ago.

#25

First, factor out 9 from the denominator, then move the 1/9 out before the integral. Now, you have x^2/(1+(x^6)/9) dx. set u=(x^3)/3 , since in the original form, u is squared. du= x^2 dx, so you are left with integral of du/(1+u^2). Substitute u (x^3/3) back into the original equation, and add the 1/9 you factored out. du/(1+u^2) is the form for tan inverse/arctan, so you are left with (1/9) tan-1(x^3/3) + C.

I'm in Calc II, and I just learned this not even 2 hours ago.

This looks pretty good, better substitution than u = x^3, makes things a bit more simple.

#26

Thank you F1rst_T and Vicd08 (awesome avatar btw). Vicd, your answer makes much more sense to me. I like the factoring out of the 9 at the beginning. THe problem is, you guys have conflicting answers, although Vicd's answer is at the back of my book. Problem solved .?

#27

Thank you F1rst_T and Vicd08 (awesome avatar btw). Vicd, your answer makes much more sense to me. I like the factoring out of the 9 at the beginning. THe problem is, you guys have conflicting answers, although Vicd's answer is at the back of my book. Problem solved .?

Yeah, he's right. My bad....sorry I wasn't more help, hah.

#28

haha nah its okay. U helped push me along. Thanks guys!

#29

42

now i realise why i failed A-level maths :P

good luck with it tho

now i realise why i failed A-level maths :P

good luck with it tho

*Last edited by nickyjohnson at Jan 27, 2009,*