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Reply 2860
Original post by Mubariz
Or wolfram alpha :ahee:

How's it going?

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True, my rival :wink:

It's going good, watching netflix, smashing out integrals and playing ps2, that's what you call a relaxing summer.

How about urs?
Original post by plusC
Hey!, thank you :smile:

That I am, if you ever need help with an indefinite integral I'm your guy


I need help :redface:

By decomposing y2+2y+1y4+2y2+1 \dfrac{y^2 + 2y + 1}{y^4 + 2y^2 + 1} into partial fractions, find

y2+2y+1y4+2y2+1dy. \displaystyle \int \dfrac{y^2 + 2y + 1}{y^4 + 2y^2 + 1} \text{d}y.

Have fun :redface:

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Reply 2862
Original post by Arieisit
I need help :redface:

By decomposing y2+2y+1y4+2y2+1 \dfrac{y^2 + 2y + 1}{y^4 + 2y^2 + 1} into partial fractions, find

y2+2y+1y4+2y2+1dy. \displaystyle \int \dfrac{y^2 + 2y + 1}{y^4 + 2y^2 + 1} \text{d}y.

Have fun :redface:

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Looks like fun! I'll give it a go :h:
Original post by plusC
Looks like fun! I'll give it a go :h:


:biggrin:

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Wazza Aaron

Original post by plusC
True, my rival :wink:

It's going good, watching netflix, smashing out integrals and playing ps2, that's what you call a relaxing summer.

How about urs?


Ps2? Hitting them vintage games yeah?

Yeah, downloading movies, trying to think of a research project, playing Xbox and waiting for my university taster weeks :ahee:

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Original post by Mubariz
Wazza Aaron



Ps2? Hitting them vintage games yeah?

Yeah, downloading movies, trying to think of a research project, playing Xbox and waiting for my university taster weeks :ahee:

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Do you say wazza in real life? :lol:

Anyway, I'll be off in a bit. Got a few things to do.

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Reply 2866

y2+2y+1y4+2y2+1=y2+1(y2+1)2+2y(y2+1)2\dfrac{y^{2}+2y+1}{y^{4}+2y^{2}+1}=\dfrac{y^{2}+1}{(y^{2}+1)^{2}}+\dfrac{2y}{(y^{2}+1)^{2}}

y2+2y+1y4+2y2+1=arctan(y)1y2+1+C\displaystyle \int \dfrac{y^{2}+2y+1}{y^{4}+2y^{2}+1}=\arctan(y)-\dfrac{1}{y^{2}+1}+C

probably not following instructions :tongue:, but this is the answer as far as I'm concerned
Original post by Mubariz
Wazza Aaron



Ps2? Hitting them vintage games yeah?

Yeah, downloading movies, trying to think of a research project, playing Xbox and waiting for my university taster weeks :ahee:

Posted from TSR Mobile

Yeah precisely, got pretty addicted to the japanese version of kingdom hearts 2, been playing it non stop really haha

This is the life :lol:
Original post by plusC
y2+2y+1y4+2y2+1=y2+1(y2+1)2+2y(y2+1)2\dfrac{y^{2}+2y+1}{y^{4}+2y^{2}+1}=\dfrac{y^{2}+1}{(y^{2}+1)^{2}}+\dfrac{2y}{(y^{2}+1)^{2}}

y2+2y+1y4+2y2+1=arctan(y)1y2+1+C\displaystyle \int \dfrac{y^{2}+2y+1}{y^{4}+2y^{2}+1}=\arctan(y)-\dfrac{1}{y^{2}+1}+C

probably not following instructions :tongue:, but this is the answer as far as I'm concerned

Yeah precisely, got pretty addicted to the japanese version of kingdom hearts 2, been playing it non stop really haha

This is the life :lol:


No, it's wrong. :no: Sorry :redface:

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I wish I could join in the maths discussion but I didn't do further maths :cry:

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Reply 2869
Original post by Arieisit
No, it's wrong. :no: Sorry :redface:

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Is the actual answer correct though? Its just I definitely know those aren't partial fractions, but why would you when you have this neat little trick :wink:
Original post by Arieisit
Do you say wazza in real life? :lol:

Anyway, I'll be off in a bit. Got a few things to do.

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Sometimes yeah.

My friends use it more than me, I rarely use it in real life tbf.

Later :wavey:

Original post by plusC
y2+2y+1y4+2y2+1=y2+1(y2+1)2+2y(y2+1)2\dfrac{y^{2}+2y+1}{y^{4}+2y^{2}+1}=\dfrac{y^{2}+1}{(y^{2}+1)^{2}}+\dfrac{2y}{(y^{2}+1)^{2}}

y2+2y+1y4+2y2+1=arctan(y)1y2+1+C\displaystyle \int \dfrac{y^{2}+2y+1}{y^{4}+2y^{2}+1}=\arctan(y)-\dfrac{1}{y^{2}+1}+C

probably not following instructions :tongue:, but this is the answer as far as I'm concerned

Yeah precisely, got pretty addicted to the japanese version of kingdom hearts 2, been playing it non stop really haha

This is the life :lol:


Na, I prefer new games, I don't like vintage games.

Meh, each to their own

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Reply 2871
Original post by Eva.Gregoria
I wish I could join in the maths discussion but I didn't do further maths :cry:

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:console:, can always self teach it :wink:
Original post by Eva.Gregoria
I wish I could join in the maths discussion but I didn't do further maths :cry:

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I haven't done it either.

Yet

Original post by plusC
:console:, can always self teach it :wink:


May aswell run at a brick wall

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Original post by plusC
Is the actual answer correct though? Its just I definitely know those aren't partial fractions, but why would you when you have this neat little trick :wink:


It is partial fractions, you have to "spot the trick" :wink:

Your answer is correct but you could've easily put it into wolfarm :tongue:

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Original post by Eva.Gregoria
I wish I could join in the maths discussion but I didn't do further maths :cry:

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I haven't done FM either :laugh:

Original post by Mubariz
I haven't done it either.

Yet



May aswell run at a brick wall

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Thought you told me you did FM in PM? :confused:

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Original post by plusC
:console:, can always self teach it :wink:


Haha I tried at the beginning of my gap year like :lol: I actually almost completed complex numbers too but I just couldn't be bothered to continue without the threat of failing exams or teachers to motivate me :redface:

Original post by Mubariz
I haven't done it either.

Yet



May aswell run at a brick wall

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It's not that difficult imo, if you're good at maths, getting A or A*, the jump shouldn't be too difficult lol.

Original post by Arieisit
I haven't done FM either :laugh:
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So then what was that beastly looking equation you posted? :frown:

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(edited 9 years ago)
Reply 2876
Original post by Arieisit
It is partial fractions, you have to "spot the trick" :wink:

Your answer is correct but you could've easily put it into wolfarm :tongue:

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Might as well do that, I'm not familiar with integration with complex conjugate roots on the denominator
Original post by Arieisit
I haven't done FM either :laugh:



Thought you told me you did FM in PM? :confused:

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I've Chosen FM.

I'm doing it the year coming with my A2's

I did my Maths AS and A2 in one year.

Kinda complicated

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Original post by Eva.Gregoria
Haha I tried at the beginning of my gap year like :lol: I actually almost completed complex numbers too but I just couldn't be bothered to continue without the threat of failing exams or teachers to motivate me :redface:



It's not that difficult imo, if you're good at maths, getting A or A*, the jump shouldn't be too difficult lol.



So then what was that beastly looking equation you posted? :frown:

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I have CAPE to thank for that. Lol. (check my profile if you're confused)

Original post by plusC
Might as well do that, I'm not familiar with integration with complex conjugate roots on the denominator


Complex conjugate roots?! Noo.
When decomposed to partial fractions

y2+2y+1y4+2y2+1=1y2+1+2y(y2+1)2\dfrac{y^2 + 2y + 1}{y^4 + 2y^2 + 1} = \dfrac{1}{y^2 + 1} + \dfrac{2y}{(y^2 + 1)^2}

Original post by Mubariz
I've Chosen FM.

I'm doing it the year coming with my A2's

I did my Maths AS and A2 in one year.

Kinda complicated

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Oh ok, I get it.



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Reply 2879
Original post by Arieisit
I have CAPE to thank for that. Lol. (check my profile if you're confused)



Complex conjugate roots?! Noo.
When decomposed to partial fractions

y2+2y+1y4+2y2+1=1y2+1+2y(y2+1)2\dfrac{y^2 + 2y + 1}{y^4 + 2y^2 + 1} = \dfrac{1}{y^2 + 1} + \dfrac{2y}{(y^2 + 1)^2}



Oh ok, I get it.



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Mate thats exactly what I did :lol: (I just didn't bother to write down the final decomposed version)