1. ## integration

i was wondering how to integrate

3/(2x - 4x^3)

im struggling to integrate this..are there steps to show this?

thanks

2. Hello,
Originally Posted by dopi
i was wondering how to integrate

3/(2x - 4x^3)

im struggling to integrate this..are there steps to show this?

thanks
$\displaystyle \frac{3}{2x-4x^3}=\frac 32 \cdot \frac{1}{x(1-2x^2)}=\frac 32 \frac{1}{x(1-\sqrt{2} x)(1+\sqrt{2} x)}$

Do you know how to use partial fraction decomposition ? Partial-Fraction Decomposition: General Techniques

3. ## still cant do it..slight mistake with equation

soory i made a slight mistake with the equation the correct one is below

$\displaystyle 3/(2x - 4x^4)$

im still struggling with breaking it down into partial fractions

4. Originally Posted by dopi
soory i made a slight mistake with the equation the correct one is below

$\displaystyle 3/(2x - 4x^4)$

im still struggling with breaking it down into partial fractions
$\displaystyle 2x-4x^4=2x(1-2x^3)=2x(1-\sqrt[3]{2}x)(1+\sqrt[3]{2}x+\sqrt[\frac{2}{3}]{x})$

5. $\displaystyle \frac{1}{x\left( 1-2x^{2} \right)}=\frac{1-2x^{2}+2x^{2}}{x\left( 1-2x^{2} \right)}=\frac{1}{x}+\frac{2x}{1-2x^{2}}.$

Ahhh, well, then having $\displaystyle \int{\frac{dx}{2x-4x^{4}}},$ put $\displaystyle x=\frac1u$ and you'll get an easy integral.