how do you integrate 1/(x^2+2)^(1/2)?

Question

thanks.

sorry i recalculatd something and it's 1/(x^2-4)^(1/2). Do i still use arcsin? cause i was under the impression arcsin is 1/(1-x^2)^(1/2)

Answer 1

1/(x^2-4)^(1/2) dx

This is in the form

du/√(u²-a²) where u =x and a = 2
∫du/√(u²-a²) = ln | u + √(u²-a²) |
so
∫1/(x^2-4)^(1/2) dx = ln | x + √(x²-4) | + C

Answer 2

assuming you mean dx, that's one of those natural log identities...
ln | x + sqrt(x^2 + 2)| + C
(the arcsin identies involve sqrt (x^2 - a^2)
Note: this is an integral of the form
1/(x^2 + a^2), where a = sqrt 2 and a^2 = 2.)
................
In regard to your edited note, that would indeed involve arcsin, with a=2.

Answer 3

integral of {1/(x^2-4)^(1/2)}

is ln {x + sqrt(x^2-4) }

Answer 4

using arcsin , it should be in the back of your calc book, i would write it out but it's hard without an equation editor