# Help me with my integral skills

1. Feb 21, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data
$$\int_0^R \! \frac{2rdr}{(r^2+x^2)^{\frac{3}{2}}}$$
$$\int_0^R \! (r^2+x^2)^{-\frac{3}{2}}}d(r^2)$$

2. Relevant equations

3. The attempt at a solution
how did you get from line 1 to 2

2. Feb 21, 2009

### rock.freak667

d(r^2)=2rdr

3. Feb 21, 2009

### -EquinoX-

oh I see... so $$d(r^2)$$ just basically means taking the derivative of $$r^2$$ with respect with r?
I am not familiar with that notation

4. Feb 21, 2009

### Tom Mattson

Staff Emeritus
Not quite. $d\left(r^2\right)$ is the differential of $r^2$ with respect to $r$.

5. Feb 21, 2009

### Unco

I bet you are! (plus or minus a twist)

Consider $$I = \int (u + a^2)^{-\frac{3}{2}} \, du.$$

Let $$u = r^2,$$ so $$du = 2r\, dr.$$ Then

$$I = \int (r^2 + a^2)^{-\frac{3}{2}} \, 2r \, dr\, \text{!}$$

6. Feb 21, 2009

### -EquinoX-

thanks for clearing that up