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

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Answer 1 · 2016-03-12T12:49:57

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

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

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