i was wondering how to integrate
3/(2x - 4x^3)
im struggling to integrate this..are there steps to show this?
thanks
Hello,
$\displaystyle \frac{3}{2x-4x^3}=\frac 32 \cdot \frac{1}{x(1-2x^2)}=\frac 32 \frac{1}{x(1-\sqrt{2} x)(1+\sqrt{2} x)}$
Do you know how to use partial fraction decomposition ? Partial-Fraction Decomposition: General Techniques
$\displaystyle \frac{1}{x\left( 1-2x^{2} \right)}=\frac{1-2x^{2}+2x^{2}}{x\left( 1-2x^{2} \right)}=\frac{1}{x}+\frac{2x}{1-2x^{2}}.$
Ahhh, well, then having $\displaystyle \int{\frac{dx}{2x-4x^{4}}},$ put $\displaystyle x=\frac1u$ and you'll get an easy integral.