$\cfrac{1}{{\sqrt {1 + 4{x^2}} }}$

: Let $I = \int {\cfrac{1}{{1 + 4{x^2}}}} dx = \cfrac{1}{2}\int {\cfrac{{dx}}{{\sqrt {\cfrac{1}{4} + {x^2}} }}}$

$= \cfrac{1}{2}\int {\cfrac{{dx}}{{\sqrt {{{\left( {\cfrac{1}{2}} \right)}^2} + {x^2}} }} = \cfrac{1}{2}\log \left| {x + \sqrt {\cfrac{1}{4} + {x^2}} } \right| + {C_1}}$

$= \cfrac{1}{2}\log \left| {\cfrac{{2x + \sqrt {1 + 4{x^2}} }}{2}} \right| + {C_1}$

$= \cfrac{1}{2}\log \left| {2x + \sqrt {1 + 4{x^2}} } \right| - \cfrac{1}{2}\log 2 + {C_1}$

$= \cfrac{1}{2}\log \left| {2x + \sqrt {1 + 4{x^2}} } \right| + C$ $\left[ {C = \cfrac{{ - 1}}{2}\log 2 + {C_1}} \right]$

Alternative solution :
Let $I = \int {\cfrac{1}{{\sqrt {1 + 4{x^2}} }}dx} = \int {\cfrac{{dx}}{{\sqrt {1 + {{\left( {2x} \right)}^2}} }}}$

Put $2x = \tan \theta$ $\Rightarrow$ $2dx = {\sec ^2}\theta d\theta$

$\therefore$ $I = \cfrac{1}{2}\int {\cfrac{{{{\sec }^2}\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}d\theta } = \cfrac{1}{2}\int {\cfrac{{{{\sec }^2}\theta }}{{\sec \theta }}d\theta } = \cfrac{1}{2}\int {\sec \theta d\theta }$

$= \cfrac{1}{2}\log \left| {\sec \theta + \tan \theta } \right| + C = \cfrac{1}{2}\log \left| {\tan \theta + \sqrt {1 + {{\tan }^2}\theta } } \right| + C$

$= \cfrac{1}{2}\log \left| {2x + \sqrt {1 + {{\left( {2x} \right)}^2}} } \right| + C = \cfrac{1}{2}\log \left| {2x + \sqrt {1 + 4{x^2}} } \right| + C$