Question

Evaluate the definite integral. I will write the equation with teh equation box

Answer 1

\[\int\limits_{1}^{\infty} dx / (x \sqrt{x^2-1})\]

Answer 2

is it du/u

Answer 3

No, not quite

Answer 4

is it a trig integral? its looks like one

Answer 5

wait, is the integral
\[ \int \frac{x}{\sqrt{x^2-1}} \ dx \]?

Answer 6

no dx is in the numerator and the numerator is x*sqrt of (x^2 -1)

Answer 7

i mean teh denominaor is x*sqrt of (....)

Answer 8

may u pplease show work to see how it gets to infinity

Answer 9

I mean it behaves like constant*1/sqrt(x-1) which does converge.

Answer 10

So use the substitution u = sqrt(x^2 - 1) and grind it through.

I'll tell you the indefinite integral is arctan( some function of x )

Answer 11

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

Now integrate that and you have arctan(u)

Therefore
\[ \int_1^\infty \frac{dx}{x\sqrt{x^2-1}} = \left[ \arctan(\sqrt{x^2-1}) \right]_1^\infty = ... \]

Answer 12

Thank You! This is a good way for me to start