Answer
$$ = - \frac{1}{2}\ln \frac{{1 + \sqrt {1 - 4{x^2}} }}{{\left| {2x} \right|}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{2x\sqrt {1 - 4{x^2}} }}} dx{\text{ can be written as }}\int {\frac{1}{{2x\sqrt {1 - {{\left( {2x} \right)}^2}} }}} dx \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& {\text{Substituting}} \cr
& {\text{ }}\int {\frac{1}{{2x\sqrt {1 - {{\left( {2x} \right)}^2}} }}} dx = \int {\frac{1}{{u\sqrt {1 - {u^2}} }}\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\frac{1}{{u\sqrt {1 - {u^2}} }}} du \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{u\sqrt {{a^2} - {u^2}} }} = - \frac{1}{a}\ln \frac{{a + \sqrt {{a^2} - {u^2}} }}{{\left| u \right|}} + } C \cr
& = \frac{1}{2}\left( { - \frac{1}{1}\ln \frac{{1 + \sqrt {1 - {u^2}} }}{{\left| u \right|}}} \right) + C \cr
& = - \frac{1}{2}\ln \frac{{1 + \sqrt {1 - {u^2}} }}{{\left| u \right|}} + C \cr
& {\text{Write in terms of }}x \cr
& = - \frac{1}{2}\ln \frac{{1 + \sqrt {1 - 4{x^2}} }}{{\left| {2x} \right|}} + C \cr} $$
