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