# Integral question

1. Nov 12, 2006

### kvanr

Just wondering if you can help me out, I have the following integral that I go to in some differential equation I came up with to figure something out here:

Integrate: dy/(4-y^0.5)

The root y screws me up.

2. Nov 12, 2006

### Hurkyl

Staff Emeritus
Then get rid of it. (With some sort of substitution)

3. Nov 12, 2006

### kvanr

Yah, how?
That's my question.

4. Nov 12, 2006

### Pyrrhus

Try

$$z^{2} = y^{0.5}$$

$$dy = 4z^3 dz$$

5. Nov 12, 2006

### kvanr

Ah yes, interesting. So then I get it in the form

$$4z^3 dz / (4 - z^2)$$

I see you can simplify the $$4 - z^2 to: (2-z)(2+z)$$, but then I tried to do partial fractions, and that also went nowhere, as there is no existing way to do a partial fraction for that that I could find?

and again I am stumped.. sorry, I'm just not getting this.

6. Nov 12, 2006

### Hurkyl

Staff Emeritus
Cyclovenom: Why not just z = y^0.5? (or z^2 = y)

This is a routine application of partial fractions. Maybe you forgot about a step? (the way I learned it... the first step)

7. Nov 12, 2006

### kvanr

$$A/(2-x) + B/(2+x) = 4x^3/(4-x^2)$$
$$A(2+x) + B(2-x) = 4x^3$$
$$(2A+2B) + x(A-B) = 4x^3 + 0x + 0$$
2A+2B = 0, so A=-B
x(A-B)=0x, so A=B

Therefore, no solution exists by the method I know.

8. Nov 12, 2006

### Hurkyl

Staff Emeritus
There's something you're supposed to do before this...

(if you can't find it in your book, the answer is: you're supposed to divide first)

9. Nov 12, 2006

### Pyrrhus

Yes, it's easier, i guess i am gettin' rusty.

10. Nov 12, 2006

### kvanr

I'm sorry I can't find it.

It's because I'm stupid.

edit: Long division.. hmm..

would it be the same as:
$$-4x + 16x / (4-x^2)$$

I checked and looks fine.. integrating now

so final answer of $$-2z^2 - 8ln(4-z^2)$$
Then just sub in root y for z^2

Last edited: Nov 12, 2006
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