help asap!!! find the integral of 1/(x^2 *sqrt(4x^2-25)) dx?

Answer 1

Find the integral of ∫1/(x²√(4x² - 25)) dx.

∫1/(x²√(4x² - 25)) dx
Let
x = (5/2)secθ
dx = (5/2)(secθ)(tanθ) dθ
x² = (25/4)sec²θ

= ∫(5/2)(secθ)(tanθ) / [(25/4)sec²θ√(4(25/4)sec²θ - 25)] dθ

= (2/5)∫(tanθ) / [secθ√(25sec²θ - 25)] dθ

= (2/5)∫(tanθ) / [(secθ)(5tanθ)] dθ

= (2/25)∫1 / (secθ) dθ

= (2/25)∫cosθ dθ

= (2/25)sinθ
__________

x = (5/2)secθ
2x/5 = secθ
5/(2x) = cosθ
sinθ = √(1 - cos²θ) = √[1 - 25/(4x²)] = √(4x² - 25) / (2x)
_____________

= (2/25)√(4x² - 25) / (2x) + C

= √(4x² - 25) / (25x) + C

Answer 2

1/(x^2 *sqrt(4x^2-25)) =1/(x^2 *2sqrt(x^2-(25/4))) letting x=2.5cosh t, dx=2.5sinhh tdt then
dx/(x^2 *sqrt(4x^2-25))= {2.5sinhh tdt}/[2.5cosh t]^2*sqrt[(2.5cosh t)^2 -25/4]
this means 1/(x^2 *sqrt(4x^2-25)) dx=dt/[2.5cosh t]^2=(4/25)tanh t +C
=(4/25)tanh[acosh x/2.5] +C