Answer
$$\frac{x}{2}\sqrt {4{x^2} - 9} - \frac{9}{4}\ln \left| {2x + \sqrt {4{x^2} - 9} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {4{x^2} - 9} dx} \cr
& u = 2x,{\text{ }}du = 2dx,{\text{ }}a = 3 \cr
& \int {\sqrt {4{x^2} - 9} dx} = \int {\sqrt {{u^2} - {a^2}} \left( {\frac{1}{2}du} \right)} \cr
& = \frac{1}{2}\int {\sqrt {{u^2} - {a^2}} du} \cr
& {\text{a matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula 68}} \cr
& \int {\sqrt {{u^2} - {a^2}} du} = \frac{u}{2}\sqrt {{u^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C \cr
& = \frac{1}{2}\int {\sqrt {{u^2} - {a^2}} du} = \frac{u}{4}\sqrt {{u^2} - {a^2}} - \frac{{{a^2}}}{4}\ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C \cr
& u = 2x,{\text{ }}a = 3 \cr
& = \frac{{2x}}{4}\sqrt {4{x^2} - 9} - \frac{{{{\left( 3 \right)}^2}}}{4}\ln \left| {2x + \sqrt {4{x^2} - 9} } \right| + C \cr
& = \frac{x}{2}\sqrt {4{x^2} - 9} - \frac{9}{4}\ln \left| {2x + \sqrt {4{x^2} - 9} } \right| + C \cr} $$
