Integral of 1 /x+ sq. root x^2-1

I'm assuming the integral is

Rationalize the denominator:

Then the integral is

For the remaining integral, substitute x = sec(t ) and dx = sec(t ) tan(t ) dt. Then over an appropriate domain, we have

Integrate by parts, taking

u = tan(t )   ==>   du = sec²(t ) dt

dv = sec(t ) tan(t ) dt   ==>   v = sec(t )

Then

Now for *this* remaining integral, integrate by parts again, taking

u = sec(t )   ==>   du = sec(t ) tan(t ) dt

dv = sec²(t ) dt   ==>   v = tan(t )

so that

To summarize, if I denotes the original integral, we have

Putting everything back in terms of x, we have

sec(t ) = x

tan(t ) = √(x ² - 1)

so that

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length