Shoutbox

Wow!  It appears that everyone is getting something different.  Here is what I did.  I used substitution and realized that the 1/x^2 can be rewritten as 1x^-2.

Integration: (x + 1/x) (sqrroot(x^2+x^-2)) dx

u = x^2+x^-2 dx
du = 2x - 2x^-1 dx
(1/2)du = x - x^-1
x- x^-1 = x + 1/x

[color=black]

Substitute (x + 1/x) with 1/2 and x^2+x^-2 with "u"

Integrate from here: 1/2 (u^(1/2))

1/2 (u^(3/2)) / (3/2)


      2u^(3/2)
        2 * 3

this equals: 2(x^2 + 1/x^2)^(3/2)
                                 6

THIS WAS CONFIRMED BY A TI-89 Titanium.  Check it yourself.  No integration by parts neccessary!

quote:

---

Originally posted by Buttah N Bred
Wow!  It appears that everyone is getting something different.  Here is what I did.  I used substitution and realized that the 1/x^2 can be rewritten as 1x^-2.

Integration: (x + 1/x) (sqrroot(x^2+x^-2)) dx

u = x^2+x^-2 dx
du = 2x - 2x^-1 dx
(1/2)du = x - x^-1
x- x^-1 = x + 1/x

[color=black]

Substitute (x + 1/x) with 1/2 and x^2+x^-2 with "u"

Integrate from here: 1/2 (u^(1/2))

1/2 (u^(3/2)) / (3/2)


      2u^(3/2)
        2 * 3

this equals: 2(x^2 + 1/x^2)^(3/2)
                                 6

THIS WAS CONFIRMED BY A TI-89 Titanium.  Check it yourself.  No integration by parts neccessary!

---

Nice math but.. did you see the date of the other posts? Months ago.

no,  I wasn't focused on the date that it was posted.  this was for anybody else who may have had this problem.