$\displaystyle

\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=

\int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}

$

i got read of one root but instead i got another one

??

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- Feb 18th 2009, 06:37 AMtransgalacticroot integral question..
$\displaystyle

\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=

\int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}

$

i got read of one root but instead i got another one

?? - Feb 18th 2009, 06:43 AMJester
- Feb 18th 2009, 09:47 AMtransgalactic
when i substituted i didnt get your expression

i got

$\displaystyle

\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)}

$ - Feb 18th 2009, 10:02 AMJester
- Feb 19th 2009, 11:58 AMtransgalactic
i cant solve this last part:

$\displaystyle

\frac{1}{3}\int\frac{-27u+2}{2u^2-u+1}

$

even when i split it:

and take t=2u^2 -u +1

?? - Feb 19th 2009, 12:31 PMrunning-gag
You should get $\displaystyle

\int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}

$

You have to split into 2 parts

$\displaystyle \int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1} = -\frac{3}{8}\int\frac{4u-1}{2u^2-u+1} - \frac{1}{8}\int\frac{1}{2u^2-u+1}$

The first part is integrable in log

The second part is integrable in Arctan after setting t=u-1/4 then $\displaystyle v = \frac{4t}{\sqrt{7}}$ - Feb 19th 2009, 11:06 PMtransgalactic
i cant get your last part

how you got this coefficients??

2A+2B+c=0

A-B+c+d=0

B=-D

A=6

i got these equation

and i cant get your coefficients

?? - Feb 20th 2009, 07:33 AMrunning-gag
$\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{1}{u} - \frac{\frac{1}{4}}{u+1} + \frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}$

I forgot the "6" as numerator ... sorry

Multiply everything above by 6 - Feb 20th 2009, 12:09 PMtransgalactic
its still does give the last part as i got it

where did my coefficientcient equations differ yours?? - Feb 21st 2009, 01:09 AMtransgalacticdid i get the coefficient equations ..
$\displaystyle

\frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\

$

$\displaystyle

A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\

$

$\displaystyle

u^3:\\

$

$\displaystyle

2A-2B+C=0\\

$

$\displaystyle

u^2:\\

$

$\displaystyle

A+B+C+D=0\\

u:\\

$

$\displaystyle

-B+D=0\\

$

1:

$\displaystyle

A=1

$ - Feb 21st 2009, 01:25 AMAir
The denominator factorises to $\displaystyle u(1+2u^3+u^2) = u(u+1)(2u^2 - u +1)$.

Thus, The fraction becomes $\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} + \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$.

You have a sign error. $\displaystyle B$ is $\displaystyle +$ve, and not $\displaystyle -$ve. Your formula produced are correct, just a sign error has made them slightly differ.

You can continue to use gaussian elimination after you've made the correction. :) - Feb 21st 2009, 01:35 AMrunning-gag
Your equations are correct.

There is a much easier way :

$\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{1}{u(u+1)(2u^2-u+1)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$

Multiply both sides by u

$\displaystyle \frac{1}{(u+1)(2u^2-u+1)} = A - \frac{Bu}{u+1} + \frac{u(Cu+D)}{2u^2-u+1}$

And set u to 0

$\displaystyle 1 = A$

Multiply both sides by u+1

$\displaystyle \frac{1}{u(2u^2-u+1)} = \frac{A(u+1)}{u} - B + \frac{(u+1)(Cu+D)}{2u^2-u+1}$

And set u to -1

$\displaystyle -\frac{1}{4} = - B$ - Feb 21st 2009, 04:03 AMtransgalactic
i tried again

http://img13.imageshack.us/img13/2839/29878774.th.gif

here is my full way till i got the coefficients

where did i go wrong

?? - Feb 21st 2009, 04:18 AMrunning-gag
C=-9 is OK

then -3+2D=0 gives D=3/2 and not 2/3 - Feb 21st 2009, 04:45 AMtransgalactic
$\displaystyle

\int \frac{-9u+\frac{3}{2}}{2u^2-u+1}

$

but then the derivative of the denominator doesnt give us the numenator

??