1. ## IntegrL of fraction

Need to integrate z/1+4z^2 +z^4

2. your fractions don't make any sense.

3. Originally Posted by Niall101
Need to integrate z/1+4z^2 +z^4

I assume ...

$\int \frac{z}{1+4z^2+z^4} \, dz$

$\frac{z}{z^4+4z^2+4 - 3} = \frac{z}{(z^2+2)^2 - 3}
$

$\int \frac{z}{(z^2+2)^2 - 3} \, dz$

$u = z^2+2$

$du = 2z \, dz$

$\frac{1}{2} \int \frac{2z}{(z^2+2)^2 - 3} \, dz
$

$\frac{1}{2} \int \frac{1}{u^2-3} \, du$

partial fractions ...

$\frac{1}{u^2-3} = \frac{1}{2\sqrt{3}} \left(\frac{1}{u-\sqrt{3}} - \frac{1}{u+\sqrt{3}}\right)$

$\frac{1}{4\sqrt{3}} \int \frac{1}{u-\sqrt{3}} - \frac{1}{u+\sqrt{3}} \, du$

$\frac{1}{4\sqrt{3}} \left[\ln(u-\sqrt{3}) - \ln(u+\sqrt{3})\right] + C$

$\frac{1}{4\sqrt{3}} \ln\left(\frac{u-\sqrt{3}}{u+\sqrt{3}}\right) + C$

$\frac{1}{4\sqrt{3}} \ln\left(\frac{z^2+2-\sqrt{3}}{z^2+2+\sqrt{3}}\right) + C$

4. Thanks so much. Makes it look easy. I was trying to integrate by parts. Could I do something similar then for integral of

1/(3Z^3 + 7)

5. Originally Posted by Niall101
Thanks so much. Makes it look easy. I was trying to integrate by parts. Could I do something similar then for integral of

1/(3Z^3 + 7)
$\int \frac{dz}{3z^3+7}$ is dangerous.
See its result:
integrate 1/(3z^3+7) - Wolfram|Alpha

But you can do it as an infinite series.

6. Originally Posted by Miss
$\int \frac{dz}{3z^3+7}$ 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

7. Originally Posted by Niall101
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
Is this a problem in Differential Equations ?

8. Originally Posted by Miss
Is this a problem in Differential Equations ?
Yes i think I have solved now tho. Thanks!

9. Originally Posted by Niall101
Yes i think I have solved now tho. Thanks!
Hmmm. Just wondering, how did you solve it?
Did you change it to a separable equation by using the suggested substitution?

10. I tried to. Actually I just have a big mess now

11. Originally Posted by Niall101
I tried to. Actually I just have a big mess now
The problem is that you will face $\int \frac{1}{7-3z^3} \, dz$.
Did try anything to solve it?

12. I tried substitution and integration by parts but I cant find the right substitutions to use.

13. Originally Posted by Niall101
I tried substitution and integration by parts but I cant find the right substitutions to use.
Am sure, its not easy. Its a dangerous integral.
Are your sure that the exponent is 3 in the original question?

14. Originally Posted by Miss
Am sure, its not easy. Its a dangerous integral.
Are your sure that the exponent is 3 in the original question?
Original Question is "By using a suitable substitution solve the differential equation"

dy/dx = (7X - 3Y)^3