[urgent] Integration - rewriting...?

Question

[urgent] Integration - rewriting...
Please visit http://www.babycat.net/private/q1b.gif

Thank you!

Answer 1

It's been a long time since I've thought about problems like this.

The part under the radical sign could be written as:

x² - 6x + 9 - 9
(x - 3)² - 9

then further massaging it:

9[1/9(x - 3)² - 1]
9[((1/3)(x - 3))² - 1]

so then

s = (1/3)(x - 3)
3s = x - 3
x = 3s + 3

dx = 3 ds

x + 1 = 3s + 4

And the denominator would be:

√[9(s² - 1)]
=3 √(s² - 1)

And the new endpoints would be:

s = (1/3)(9 - 3) = 2

and

s = (1/3)(12 - 3) = 3

So altogether it would be:

3
⌠ (3s + 4) (3 ds)
---------------------------
⌡ 3 √(s² - 1)
2

3
⌠ (3s + 4) ds
------------------
⌡ √(s² - 1)
2

Dumb mistake the first time made it much uglier.

Answer 2

Well, here is a hint.

the x^2-6x can be written as (x-3)^2-9. Then do a variable substitution, say, y=x-3 so that the denominator becomes
y^2-9. Another substitution will give you the form w^2-1.

Good luck.

Answer 3

x^2 - 6x = x^2 - 2.x.3 + 3^2 - 3^2 = (x - 3)^2 - (3)^2

So Your form
(mx +c)/sqrt[x^2 - a^2)]
k = 1

Answer 4

x^2-6x = (x-3)^2-3^2 =3^2{[(x-3)/3]^2 - 1}