# Help me with my integral skills

#### -EquinoX-

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

2. Homework Equations

3. The Attempt at a Solution
how did you get from line 1 to 2

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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

#### Tom Mattson

Staff Emeritus
Gold Member
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

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