`int_sqrt(3)^3 1/ (xsqrt(4x^2-9)) dx` Evaluate the definite integral

Make the substitution  `u = sqrt(4x^2 - 9),` then  `du = (4x)/sqrt(4x^2 - 9) dx.` Inversely, `dx =sqrt(4x^2 - 9)/(4x) du = u/(4x) du`  and  `4x^2 = u^2 + 9.` The limits of integration become from  `sqrt(3)`  to  `3sqrt(3).`

The indefinite integral becomes

`int u/(4 u x^2) du = int (du)/(u^2 + 9) = 1/3 arctan(u/3) + C,`

where `C` is an arbitrary constant.

Thus the definite integral is  `1/3 (arctan(sqrt(3)) - arctan(1/sqrt(3))) = 1/3 (pi/3 - pi/6) = pi/18.`