Find x and dx using trigonometric substitution of

(integral) square root of 4x^2 - 9 divide by x

x = 2/3 sec x
dx = 2/3 tan x dx

Is this right so far?

∫√(4x^2-9) dx

2x = 3secθ, so
4x^2-9 = 9sec^2θ-9 = 9tan^2θ
x = 2/3 secθ, so
dx = 2/3 secθ tanθ

and you integral becomes

∫3tanθ (2/3 secθ tanθ) dθ
= 2∫secθ tan^2θ dθ
= 2∫(sec^3θ - secθ) dθ

use integration by parts twice for sec^3θ.

rats. x = 3/2 secθ, so adjust the constant in the following steps.

when you're all done, you can check your answer by typing in the integral at wolframalpha.com

ok thank you very much. Happy New Year

Yes, your substitution is correct so far.

To verify if it is correct, you can differentiate x = 2/3 sec(x) to find dx.

Taking the derivative of both sides with respect to x, we have:

d/dx (x) = d/dx (2/3 sec(x))

1 = 2/3 sec(x) tan(x)

Multiplying both sides by 3/2 sec(x), we get:

3/2 sec(x) = 2/3 tan(x)

So dx = 2/3 tan(x) dx. Your substitution is therefore confirmed to be correct.