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

Let $I = \int {\sqrt {{x^2} + 4x + 1} } dx = \int {\sqrt {({x^2} + 4x + 4) - 3} } dx$

$= \int {\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} dx}$

$= \cfrac{{x + 2}}{2}\sqrt {{{\left( {x + 2} \right)}^2} - 3} - \cfrac{3}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{{\left( {x + 2} \right)}^2} - 3} } \right| + C$

$= \cfrac{{x + 2}}{2}\sqrt {{x^2} + 4x + 1} - \cfrac{3}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 1} } \right| + C$