Question

Here is a problem that wolfram won't evaluate that someone had a question about.
\[int_{}^{} sqrt{1+\frac{1}{4} \cdot frac{x^2 (x-2)^2}{(1-x)^2}} dx\]

Answer 1

\[\int\limits_{}^{} \sqrt{1+\frac{1}{4} \cdot \frac{x^2 (x-2)^2}{(1-x)^2}} dx \]

Answer 2

I honestly don't see a way by elementary ways.

Answer 3

JamesJ can do it!

Answer 4

Let's see if my calc evaluate it :P

Answer 5

Otherwise I will try sympy

Answer 6

breaking it into partial fractions is a possibility

Answer 7

It doesn't come out to anything simple at all

Answer 8

partial fractions then substitution

Answer 9

Ok, the integrand is equal to
\[ \frac{1}{2} \sqrt{ \frac{ 4(1-x)^2 + x^2(x-2)^2 }{ (1-x)^2 } } = \frac{1}{2} \sqrt{ \frac{ (x^2 - 2x + 2)^2}{(1-x)^2} } \] \[ = \frac{1}{2} \frac{ x^2 - 2x + 2}{1-x} = -x - \frac{1}{x-1} + 1\]

Answer 10

Well, someone said james could do it and he did it

Answer 11

why didn't i see that?

Answer 12

I'm just happy to find another example of Wolfram Alpha defeated by pencil and paper. :-)

Answer 13

good job
another point for the human team

Answer 14

pencil and paper > modern CAS

Answer 15

TI-Nspire CAS defeated too

Answer 16

Apparently James is traditionalist! ;D

Answer 17

I couldn't figure out how to use sympy :-P

Answer 18

**Correction: I dropped the half in the last expression.

Answer 19

that's cool who cares about the constant multiple we put it in a form easy to integrate
well you did anyways

Answer 20

The factorization of the numerator isn't obvious. Don't feel bad.

But I do think this problem is a bit silly. What was the motivation for that integrand in that form? I'd bet a nickel the integrand was constructed from the simplified form as a puzzle, rather than being motivated by any other problem in mathematics or the sciences.

Answer 21

lol i made a puzzle by accident

Answer 22

the original problem was find the length of
\[y=\frac{x}{2}-\frac{x^2}{4}+\frac{1}{2}\ln(1-x) \text{ on interval } [0, .5]\]
so i found y'
which is
\[y'=\frac{1}{2}(1-x-\frac{1}{x-1})\]
then i wrote (y')^2 as
\[(y')^2=\frac{1}{4} \cdot \frac{x^2(x-2)^2}{(1-x)^2}\]
which made it uglier than it should had been

Answer 23

wasn't thinking

Answer 24

No, I owe you a nickel. This did come from a real problem.

Answer 25

What do you mean?

Answer 26

I mean the integral came from a real problem, vs. just being a slightly annoying puzzle.

Answer 27

Ok. lol

Answer 28

so what was your trick for factoring the numerator?

Answer 29

he used... wolfram!

Answer 30

no james doesn't use wolfram he doesn't believe in it lol

Answer 31

right

Answer 32

I knew I wanted a perfect square to get rid of the square trot. So I went looking for it. As ever with three things squared, I think it's easier to find it as
\[ ((a+b) + c)^2 \]
vs as
\[ (a+b+c)^2 \]
So after I realized it wasn't (a+b)^2, I went looking for things in that form.

Answer 33

*square root

Answer 34

JamesJ - this is the 2nd such problem that wolfram couldn't deal with. should these be reported to the wolfram team so that they can improve their algorithms?

Answer 35

Geez asnaneer. Just send me to the glue factory already. ;-)

Answer 36

the integral algorithm is probably complicated :-D

Answer 37

James thanks you are the coolest i think i learned a little

Answer 38

:-) - just trying to improve the mathematical community - thats all...

Answer 39

I bet that on JamesJ's copy of the monograph on the Riemann Hypothesis is a note on the margin saying : "I have discovered a truly marvellous proof of this, which this margin is too narrow to contain."

Answer 40

BTW JamesJ - what type of glue do you prefer? :-D

Answer 41

I'm sure Wolfram could find it, but it has to make decisions about how long to run certain algorithms. With a tweak of its "time out" parameters (or @agd, tree pruning heuristic), it would find this solution.

Answer 42

http://www.wolframalpha.com/input/?i=integrate+Sqrt%5B1+%2B+1%2F4+%28x%5E2+%28x+-+2%29%5E2%29%2F%281+-+x%29%5E2%5D

Answer 43

the wolfram solution is not as elegant as JamesJ's solution though.

I think it's an interesting exercise to do a little research on how those symbolic integration/factoring stuff work

Answer 44

I dread to think how it found that solution, because it suggests quite a tortured process. But the real problem is once it did, it stopped and it didn't try and simplify the answer, which clearly it could have.

But in any case, it looks like it didn't use the procedure I did: try and find a perfect square and rid of the square root asap. Anyway, time to celebrate by making a cup of tea. ;-)

Answer 45

I agree - plus the wolfram solution doesn't really improve your understanding of the problem.

Answer 46

lol

Answer 47

let's see what that function returns when I try to evaluate myininaya's integral

Answer 48

still running :-P

Answer 49

http://ideone.com/9Cxc3 here's the integral method again

Answer 50

I think we should check in 50 minutes for the output of the integral method. :-D

Answer 51

\[#
______________________
/ 2 2
/ 0.25*x *(x - 2)
/ ---------------- + 1
/ 2
\/ (-x + 1) \]
was what I gave it,

Answer 52

It couldn't do it. It returned :\

/
|
| _________________
| / 2 2
| / x *(x - 2)
| / ----------- + 1 dx
| / 2
| \/ 4*(-x + 1)
|
/