# root integral question..

• Feb 18th 2009, 06:37 AM
transgalactic
root 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 AM
Jester
Quote:

Originally Posted by transgalactic
$\displaystyle \int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}= \int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}$
i got read of one root but instead i got another one
??

If you let $\displaystyle x = u^6$ then

$\displaystyle \int \frac{6\,du}{u(u+1)(2u^2-u+1)}$
• Feb 18th 2009, 09:47 AM
transgalactic
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 AM
Jester
Quote:

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

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

$\displaystyle 2u^3+u^2+ 1 = (u+1)(2u^2-u+1)$
• Feb 19th 2009, 11:58 AM
transgalactic
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 PM
running-gag
Quote:

Originally Posted by transgalactic
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

??

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 PM
transgalactic
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 AM
running-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 PM
transgalactic
its still does give the last part as i got it

where did my coefficientcient equations differ yours??
• Feb 21st 2009, 01:09 AM
transgalactic
did 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 AM
Air
Quote:

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

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 AM
running-gag

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 AM
transgalactic
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 AM
running-gag
C=-9 is OK
then -3+2D=0 gives D=3/2 and not 2/3
• Feb 21st 2009, 04:45 AM
transgalactic
$\displaystyle \int \frac{-9u+\frac{3}{2}}{2u^2-u+1}$

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