# Can't Get Integral To Come Out Right

1. Nov 4, 2012

### Meadman23

1. The problem statement, all variables and given/known data

What is ∫r^3/(r^2 + h^2) ^ (3/2)

2. Relevant equations

uv - ∫v du

3. The attempt at a solution

Using Maxima gives me (r^2 + 2h^2) / sqrt(r^2 + h^2), but when I solve it using integration by parts, I get (3r^2 + 4h^2) / sqrt (r^2 + h^2).

I do:

dv = r(r^2 + h^2) ^ (-3/2)
v = -1/sqrt(r^2 + h^2)

u = r^2
du = 2r dr

Then I plug it into the int. by parts formula.

uv - ∫v du = -r^2/ sqrt (r^2 + h^2) - ∫ - 2r dr / (r^2 + h^2)

And then I integrate once more and I'm left with:

-r^2 / sqrt (r^2 + h^2) + 4*sqrt (r^2 + h^2)

I know that this should work out almost exactly like example 7 from this page, but it just won't. What's going on??

2. Nov 4, 2012

### LCKurtz

Try a straight u substitution instead of parts. Write your numerator as $r^2\cdot r\, dr$ and let $u=r^2+h^2$.

3. Nov 4, 2012

### Meadman23

Doesn't seem to work.

Letting u = r ^2 + h^2 takes care of r dr but not r^2.

I even tried setting r^2 = u - h^2 , but that doesn't get me anywhere either.

4. Nov 4, 2012

### SammyS

Staff Emeritus
That should (and does) work !

Break the integrand into the sum/difference of two fractions.

5. Nov 4, 2012

### LCKurtz

Sure it does. Put $u - h^2$ in for that $r^2$.

6. Nov 5, 2012

### Meadman23

Okay, thanks guys. I split the fraction up into two.

∫u/(2*u^(3/2)) - ∫h^2/(2*u^(3/2)

I wasn't able to get it to come out right until I considered the h^2 as a constant and factored it out along with the 1/2.

1/2∫u/(u^(3/2)) - h^2/2 ∫1/(2*u^(3/2)

Why can I consider h^2 as a constant here? Also, why did integration by parts provide the wrong answer for me??

7. Nov 5, 2012

### SammyS

Staff Emeritus
Well, h is not a function of r, is it ?

8. Nov 5, 2012

### Meadman23

I mean, h^2 = u - r^2 which means to me that the value of h^2 is dependent on the value of r. How could it not be?

9. Nov 5, 2012

### uart

No you're confusing yourself. u and r^2 are both functions of r, but their difference is not.

Consider this trivial example. Let $y= 1 - x^2$. Now y is a function of x, obviously. But x^2 is also a function of x.

So if we rearrange this to get $1 = y - x^2$, does this mean that "1" is a function of x? No, certainly not.

Last edited: Nov 5, 2012
10. Nov 5, 2012

### Meadman23

I might be reaching here but let's see.

I have:

u = h^2 + r^2

and

r^2

u - r^2 = (h^2 + r^2) - r^2 = h^2

Thus, r^2 vanishes and h^2 is no longer of a function of r. Thus, I can factor it out of my integral.

Am I seeing this right?

11. Nov 5, 2012

### dextercioby

Hmm, let $r = h \sinh t$ as a change of variable.

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