Question

How do you integrate `int 1/sqrt(4x^2-12x+9)` using trigonometric substitution?

Answer 1

`int(1/(2x-3))dx=1/2ln(2x-3)+C`

Explanation:

denominator is a perfect square
`((2x)^2-2*2x*3+3^2)=(2x-3)^2`
hence
integral
`=int(1/(2x-3))dx=1/2ln(2x-3)+C`