please help me finish my calc hw: how do i find the integral of (x)/(square root of x^2-1) dx. Thank you!?

Question

... bunch of angels
thank you! : )

Answer 1

I would make a substitution. For instance,
let u = x^2 -1
du/dx = 2x
therefore du = 2x dx

int [ x / (sqrt {x^2 -1}) dx] = (1/2) int [ 2x/ (sqtr {x^2-1})dx]

Now, if we rewrite all this in index form, we get:

(1/2) int [ (x^2 -1)^(-1/2) * 2x dx]

Funnily enough, du = 2x dx AND x^2 -1 = u. So if we make these two substitutions into our integral, we get:

(1/2) int [ u^(-1/2) du] = (1/2) [ 2u^(1/2)]
=u^(1/2)

Now, as u = x^2 -1, we get:

u^(1/2) = (x^2-1)^(1/2)

Therefore, your integral give the Square root of (x^2 -1) + C (Dont forget the constant fo integration! Always add the + C - markers have been known to deduct for this mistake!)

Answer 2

Let U= x^2-1. Then dU= 2xdx. Then your integral converts to 1/2 dU/ sqrt(U). You should be able to handle this and back-substitute.

Answer 3

∫x/(x^2-1)^(1/2) dx

let x^2 - 1 = y^2

2x dx = 2y dy

x dx = y dy

∫x/(x^2-1)^(1/2) dx =

∫y dy/ y = ∫dy = y + c

= sqrt(x^2-1) + c