Question

Help me!!!!

Answer 1

\[ I _{n}= \int\limits_{0}^{1} \frac{ 1 }{\left( 1+x ^{4} \right)^{n}}dx \]

Answer 2

By considering
\[ \frac{ x }{ \left( 1+x ^{4} \right)}^{n} \prime\]
Show that
\[4n I _{n+1}=\frac{ 1 }{ 2^{n} }+\left( 4n-1 \right)I _{n}\]

Answer 3

sorry, the "n" in the second post should be on the denominator

Answer 4

@PeterPan Hey there! can you give me a hand on this? My math homework is really tough!

Answer 5

http://tinyurl.com/k4sw46z

Answer 6

^.^
but I'll see what I can do

Answer 7

hahaha, such a helpful hand!

Answer 8

what did they mean by consider this
\[\frac{ x }{ \left( 1+x ^{4} \right)^n} ^{ \ \ \prime}\]

Answer 9

Meaning this?
\[\Large \frac{(1+x^4)^n-4nx^2(1+x^4)^{n-1}}{(1+x^4)^{2n}}\]

Answer 10

Yes, I am supposed to derive the formula under " show that " using that

Answer 11

\[\Large \frac{1+x^4-4nx^2}{(1+x^4)^{n+1}}\]

Answer 12

Lousy grown-ups and their maths problems

Answer 13

You're right! Further Math SUCKS!

Answer 14

I'm always right! >:)

Answer 15

I might be right in saying I don't know how to do this, though :(

Answer 16

but maybe you have made a mistake, the 4nx^2 should be 4nx^4

Answer 17

I beg your pardon? >:(

Answer 18

:)
Let me have a look at it

Answer 19

Okay, you're right :D
\[\Large \frac{1+x^4-4nx^4}{(1+x^4)^{n+1}}\]

Answer 20

But what should we do next?

Answer 21

Go on an adventure :D

Answer 22

Or... Just integrate this XD
THAT's an adventure :)

Answer 23

\[\Large \int\limits_0^1\frac{1+(1-4n)x^4}{(1+x^4)^{n+1}}dx\]

Answer 24

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20-%20Further%20(9231)/9231_w08_ms_1.pdf

the answer to this question is on the page 7 of the document above. But it is overly simplified

Answer 25

Must have made a mistake in differentiating :)
Could we have another try?

Answer 26

Sorry, but it's already midnight in my city, so I have to go to bed. See you tomorrow, we'll continue then

Answer 27

No, I probably won't be back... come on, @caozeyuan stay a little longer, please? :)

Answer 28

OK, I can see the CIE people used product rule so let's try that

Answer 29

Wait... I think it has something to do with the second line...

Answer 30

What is the second line? My second post or something else?

Answer 31

Okay, it all makes sense, now... a little clever manipulation was all that's needed...

Answer 32

Let's start with this :D
\[\Large \frac{1+x^4-4nx^4}{(1+x^4)^{n+1}}\]

Answer 33

It's equal to
\[\large \frac{1}{(1+x^4)^{n}}-\frac{4nx^4}{(1+x^4)^{n+1}}\] right?

Answer 34

Yes, I got it

Answer 35

I am killed by this question!

Answer 36

This can be written as
\[\Large \frac1{(1+x^4)^{n+1}}\left[1+x^4 -4nx^4 \right]\]

Answer 37

Ok, that's easy

Answer 38

Yes... Now
\[\Large \frac1{(1+x^4)^{n+1}}\left[1+x^4\color{green}{-4n} -4nx^4\color{green}{+4n} \right]\]

Answer 39

And we have \[\Large \frac1{(1+x^4)^{n+1}}\left[(1-4n)(1+x^4)+4n \right]\]

Answer 40

\[\Large = \frac{1-4n}{(1+x^4)^n}+\frac{4n}{(1+x^4)^{n+1}}\]

Answer 41

Now just integrate it ^.^\[\Large \int\limits_0^1\left[ \frac{1-4n}{(1+x^4)^n}+\frac{4n}{(1+x^4)^{n+1}}\right]dx\]

Answer 42

Not forgetting of course, that this was the derivative of \[\frac{ x }{ \left( 1+x ^{4} \right)^n} \] and therefore that^ is an antiderivative of that thingy over there. By the Fundamental Theorem of Calculus...
\[\Large \int\limits_0^1\left[ \frac{1-4n}{(1+x^4)^n}+\frac{4n}{(1+x^4)^{n+1}}\right]dx=\left.\frac{ x }{ \left( 1+x ^{4} \right)^n}\right]_0^1\]

Answer 43

\[\Large \int\limits_0^1\left[ \frac{1-4n}{(1+x^4)^n}+\frac{4n}{(1+x^4)^{n+1}}\right]dx=\frac1{2^n}\]

Answer 44

\[\Large \Large (1-4n)\int\limits_0^1 \frac{1}{(1+x^4)^n}dx+ 4n\int\limits_0^1\frac{1}{(1+x^4)^{n+1}}dx=\frac1{2^n}\]

Answer 45

\[\Large \Large -(4n-1)I_n+ 4nI_{n+1}=\frac1{2^n}\]

Answer 46

And you already left... :(

I will not forget this >:(

Answer 47

you've thrown a bomb at my head!