Need to integrate z/1+4z^2 +z^4
Please help. thanks!
I assume ...
$\displaystyle \int \frac{z}{1+4z^2+z^4} \, dz$
$\displaystyle \frac{z}{z^4+4z^2+4 - 3} = \frac{z}{(z^2+2)^2 - 3}
$
$\displaystyle \int \frac{z}{(z^2+2)^2 - 3} \, dz$
$\displaystyle u = z^2+2$
$\displaystyle du = 2z \, dz$
$\displaystyle \frac{1}{2} \int \frac{2z}{(z^2+2)^2 - 3} \, dz
$
$\displaystyle \frac{1}{2} \int \frac{1}{u^2-3} \, du$
partial fractions ...
$\displaystyle \frac{1}{u^2-3} = \frac{1}{2\sqrt{3}} \left(\frac{1}{u-\sqrt{3}} - \frac{1}{u+\sqrt{3}}\right)$
$\displaystyle \frac{1}{4\sqrt{3}} \int \frac{1}{u-\sqrt{3}} - \frac{1}{u+\sqrt{3}} \, du$
$\displaystyle \frac{1}{4\sqrt{3}} \left[\ln(u-\sqrt{3}) - \ln(u+\sqrt{3})\right] + C$
$\displaystyle \frac{1}{4\sqrt{3}} \ln\left(\frac{u-\sqrt{3}}{u+\sqrt{3}}\right) + C$
$\displaystyle \frac{1}{4\sqrt{3}} \ln\left(\frac{z^2+2-\sqrt{3}}{z^2+2+\sqrt{3}}\right) + C$
$\displaystyle \int \frac{dz}{3z^3+7}$ is dangerous.
See its result:
integrate 1/(3z^3+7) - Wolfram|Alpha
But you can do it as an infinite series.
Oh thats nasty. I dont think I should know how to do anything like that. Im trying to solve dy/dx = (7X - 3Y)^3
And i used the substitution Z = 7X - 3Y
and i come up with that integral
Edit: I ended up with A SIMPLE ODE and i was treating it as variable seperable I think this was the problem