The value of integral ∫ √(4 x^2+9) d x is

$I=\int \sqrt{4 x^2+9} d x$

$=\int \sqrt{(2 x)^2+3^2} d x$

$\int \sqrt{y^2+3^2} \frac{d y}{2}$

let y = 2x     ......(1)

⇒  $d y=2 d x $

$\Rightarrow \frac{d y}{2}=d x$

so expanding

$\frac{1}{2}[\frac{y}{2}\sqrt{y^2+3^2} + \frac{3^2}{2}log |y+\sqrt{y^2+3^2}|]+C$

since $|\int \sqrt{x^2+a^2} dx = \frac{x}{2} \sqrt{x^2+ a^2} + \frac{a^2}{2} log |x^2+9^2| + C|$

from (1) reapplying of x in expression

$\frac{1}{2}[\frac{2x}{2} \sqrt{(2x)^2 + 3^2} + \frac{3^2}{2} log 4x + \sqrt{(2x)^2+3^2}]+ C$

$=\frac{x}{2} \sqrt{4 x^2+9}+\frac{9}{4} \log [\left(2 x+\sqrt{4 x^2+9}\right)]+C$