root integral question..

**transgalactic** · February 18th 2009, 06:37 AM

$ \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
??

**Danny** · February 18th 2009, 06:43 AM

Originally Posted by transgalactic

$ \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
??

If you let $x = u^6$ then

$\int \frac{6\,du}{u(u+1)(2u^2-u+1)}$

**transgalactic** · February 18th 2009, 09:47 AM

when i substituted i didnt get your expression
i got
$ \int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)} $

**Danny** · February 18th 2009, 10:02 AM

Originally Posted by transgalactic

when i substituted i didnt get your expression
i got
$ \int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{6u^5du}{u^6(1+2u^3+u^2)} $

I'm sure you see the $u^5$ cancels and that if $f(u) = 2u^3+u^2+ 1$ then $f(-1) = 0$ so $u+1$ is a factor and

$2u^3+u^2+ 1 = (u+1)(2u^2-u+1)$

**transgalactic** · February 19th 2009, 11:58 AM

i cant solve this last part:
$ \frac{1}{3}\int\frac{-27u+2}{2u^2-u+1} $

even when i split it:
and take t=2u^2 -u +1

??

**running-gag** · February 19th 2009, 12:31 PM

Originally Posted by transgalactic

i cant solve this last part:
$ \frac{1}{3}\int\frac{-27u+2}{2u^2-u+1} $

even when i split it:
and take t=2u^2 -u +1

??

You should get $ \int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1} $

You have to split into 2 parts

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

The first part is integrable in log
The second part is integrable in Arctan after setting t=u-1/4 then $v = \frac{4t}{\sqrt{7}}$

**transgalactic** · February 19th 2009, 11:06 PM

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
??

**running-gag** · February 20th 2009, 07:33 AM

$\frac{1}{u(1+2u^3+u^2)} = \frac{1}{u} - \frac{\frac{1}{4}}{u+1} + \frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}$

I forgot the "6" as numerator ... sorry
Multiply everything above by 6

**transgalactic** · February 20th 2009, 12:09 PM

its still does give the last part as i got it

where did my coefficientcient equations differ yours??

**transgalactic** · February 21st 2009, 01:09 AM

$ \frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\ $
$ A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\ $
$ u^3:\\ $
$ 2A-2B+C=0\\ $
$ u^2:\\ $
$ A+B+C+D=0\\ u:\\ $
$ -B+D=0\\ $
1:
$ A=1 $

**Air** · February 21st 2009, 01:25 AM

Originally Posted by transgalactic

$ \frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\ $
$ A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\ $
$ u^3:\\ $
$ 2A-2B+C=0\\ $
$ u^2:\\ $
$ A+B+C+D=0\\ u:\\ $
$ -B+D=0\\ $
1:
$ A=1 $

The denominator factorises to $u(1+2u^3+u^2) = u(u+1)(2u^2 - u +1)$ .

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

You have a sign error. $B$ is $+$ ve, and not $-$ 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.

**running-gag** · February 21st 2009, 01:35 AM

Your equations are correct.

There is a much easier way :

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

Multiply both sides by u

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

And set u to 0
$1 = A$

Multiply both sides by u+1
$\frac{1}{u(2u^2-u+1)} = \frac{A(u+1)}{u} - B + \frac{(u+1)(Cu+D)}{2u^2-u+1}$

And set u to -1
$-\frac{1}{4} = - B$

**transgalactic** · February 21st 2009, 04:03 AM

i tried again

here is my full way till i got the coefficients
where did i go wrong
??

**running-gag** · February 21st 2009, 04:18 AM

C=-9 is OK
then -3+2D=0 gives D=3/2 and not 2/3

**transgalactic** · February 21st 2009, 04:45 AM

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

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

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