integral of dx/square root(x^2-64)

Question

My answer is:

ln (x/8 + square root(x^2-64)/8) +c

is this right?

Answer 1

not even close :-(

let x = 8secu
dx = 8secu*tanu du
x^2-64 = 64sec^2 u - 64 = 64tan^2 u

dx/√(x^2-64) = 8*secu*tanu/8tanu = secu

since ∫secu = ln|secu+tanu|, we have

Recall that x = 8secu, so
secu = x/8
tanu = √(x^2/64 - 1) = 1/8 √(x^2-64)
∫secu = ln|secu+tanu| = ln|x/8 + 1/8√(x^2-64)| + c = ln|(x + √(x^2-64))/8| + c
Now, ln(n/8) = ln(n) - ln(8), so we can fold -ln(8) into c, giving us

ln|x + √(x^2-64)| + C

Answer 2

hmm. upon rereading your answer, you are in fact correct.!!