# I Griffiths problem 2.7 Messy Integral

1. Sep 7, 2016

### Waxterzz

Hi all,

"Integral can be done by partial fractions - or look it up" So second line, that's what I want to do.

How to deal with this? What substitution can I use? Never encountered partial fractions with non-integer exponents.

Someone give me a tip?

2. Sep 7, 2016

I do think the author must have meant integration by parts. The term without a $u$ in the numerator that has a $(A+Bu)^{3/2}$ in the denominator is a straightforward integration and the $u \, du/(A+Bu)^{3/2}$ can be readily integrated by parts.

3. Sep 7, 2016

### zaidalyafey

You can directly use integration by parts.

4. Sep 7, 2016

### Waxterzz

I get something extremely messy. Here is the first part, the second part with integration by parts,I'll do it again tomorrow.

So the first term

integral (z du / (R² + z² - 2 Rz u) ^3/2 )

d(R² + z² - 2 Rz u) = -2 R z du

The integral becomes

-1/2R * integral ( d(R² + z² - 2 Rz u) / (R² + z² - 2 Rz u) ^3/2) )

Then I get

1/R * (1/ (R² + z² -2Rzu) ^ 1/2)

The second part I'm going to do again tomorrow, and I'm going to try to put it in latex (have to learn latex quick) because posting like this is unreadable. But the first part is correct right?

5. Sep 7, 2016

### zaidalyafey

You can differentiate z - Ru and integrate ( du / (R² + z² - 2 Rz u)^3/2)

6. Sep 7, 2016

### mathman

From first to second line is simply the substitution given. The integration is from second line to third.

7. Sep 10, 2016

### Waxterzz

Ok, so partial integration. integral (fdg) = fg - integral (gdf)

f = z-Ru
dg = du / (R² +z² -2zRu) ^ 3/2

For g I get
1 / ( (Rz) * (R² + z² -2*R*z*u)^1/2)

fg = (z-Ru) / ( (Rz) * (R² + z² -2*R*z*u)^1/2)

now the integral part - integral ( g d f)

df = -R du

So

The integral to solve

Rdu / ((Rz) * (R² + z² -2*R*z*u)^1/2))

Becomes

1/z * (2/-2*R*z) * (R² + z² -2*R*z*u)^1/2

Combining two terms, and rearranging

I get solution

DEAR GOD

hhahaha

(Okay I really need to learn latex now haha)

Thank u also.