How to Integrate (r^2 + a^2)^-3/2 using u-substitution

[-EquinoX-]

Homework Statement

$$\int_0^R \! \frac{2rdr}{(r^2+x^2)^{\frac{3}{2}}}$$
$$\int_0^R \! (r^2+x^2)^{-\frac{3}{2}}}d(r^2)$$

Homework Equations

The Attempt at a Solution

how did you get from line 1 to 2

 

[rock.freak667]

d(r^2)=2rdr

 

[-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

 

[quantumdude]

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

 

[Unco]

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

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

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

 

[-EquinoX-]

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

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

thanks for clearing that up