Integral - partial fractions?

• Sep 17th 2008, 06:11 PM
symstar
Integral - partial fractions?
$\displaystyle \int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$

$\displaystyle u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$

$\displaystyle =\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$\displaystyle =3\int\frac{du}{u^2(2u+1)(2u-3)}$

Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.
• Sep 17th 2008, 06:14 PM
Jhevon
Quote:

Originally Posted by symstar
$\displaystyle \int \frac{dx}{4x^{2/3}-4x^{1/3}-3}$

$\displaystyle u=x^{1/3} \text{ }du=\tfrac{1}{3}x^{-2/3} \text{ }x=u^3$

$\displaystyle =\int\frac{x^{-2/3}*3du}{(2u+1)(2u-3)}$
$\displaystyle =3\int\frac{du}{u^2(2u+1)(2u-3)}$

Am I going about this the right way? I ask because when I use partial fractions, I am getting 2 different values for A.

you are ok so far

so you must be making a mistake with the partial fractions part. what did you get?
• Sep 17th 2008, 06:53 PM
Krizalid
Quote:

Originally Posted by symstar

$\displaystyle \int\frac{du}{u^2(2u+1)(2u-3)}$

Put $\displaystyle u=\frac1z$ and your integral becomes $\displaystyle -\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence

\displaystyle \begin{aligned} -\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\ & =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\ & =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\ \end{aligned}

Things should be easy from there.
• Sep 17th 2008, 06:59 PM
Jhevon
Quote:

Originally Posted by Krizalid
Put $\displaystyle u=\frac1z$ and your integral becomes $\displaystyle -\int{\frac{z^{2}}{(2+z)( 2-3z)}\,dz},$ hence

\displaystyle \begin{aligned} -\frac{z^{2}}{(2+z)(2-3z)}&=\frac{(2+z)(2-z)-4}{(2+z)(2-3z)} \\ & =\frac{2-z}{2-3z}-\frac{4}{(2+z)(2-3z)}=\frac{2-z}{2-3z}-\frac{3(2+z)+(2-3z)}{2(2+z)(2-3z)} \\ & =\frac{2-z}{2-3z}-\frac{1}{2}\left( \frac{3}{2-3z}+\frac{1}{2+z} \right). \\ \end{aligned}

Things should be easy from there.

i know what you're thinking, symstar, "how does he do it?!" (Cool)
• Sep 17th 2008, 07:10 PM
symstar
So working with the partial fractions:

$\displaystyle \frac{1}{u^2(2u+1)(2u-3)}=\frac{A}{u^2}+\frac{B}{(2u+1)}+\frac{C}{(2u-3}$
$\displaystyle 1=A(4u^2-4u-3)+B(2u^3-3u^2)+C(2u^3+u^2)$
$\displaystyle 2B+2C=0$
$\displaystyle 4A-3B+C=0$
$\displaystyle -4A=0$
$\displaystyle -3A=1$

You can see the 2 values for A at the bottom. What should I do?
• Sep 17th 2008, 07:16 PM
Soroban
Hello, symstar!

I don't agree . . .

Quote:

$\displaystyle \int \frac{dx}{4x^{\frac{2}{3}}-4x^{\frac{1}{3}}-3}$

Let: $\displaystyle u \:=\:x^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:u^3\quad\Rightarrow\quad dx \:=\:3u^2\,du$

Substitute: .$\displaystyle \int \frac{3u^2\,du}{4u^2 - 4u - 3} \;=\;\frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$

. . $\displaystyle = \;\frac{3}{4}\int\left[1 + \frac{4u+3}{4u^2-4u-3}\right]\,du \;=\;\frac{3}{4}\int\left[1 + \frac{4u+3}{(2u-3)(2u+1)}\right]\,du$

Partial Fractions: . $\displaystyle \frac{3}{4}\int\left[1 + \frac{\frac{9}{4}}{2u-3} - \frac{\frac{1}{4}}{2u+1}\right]\,du \quad\hdots\text{ etc.}$

• Sep 17th 2008, 07:39 PM
symstar
$\displaystyle \frac{3}{4}\int \frac{u^2\,du}{u^2-u-\frac{3}{4}} \;=\;\frac{3}{4}\int\left[1 + \frac{u+\frac{3}{4}}{u^2-u-\frac{3}{4}}\right]\,du$

I'm not quite sure what exactly happened in this step, could you explain please?