Answer
$$\frac{\pi }{{6\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{x}{{\sqrt {4 - 3{x^4}} }}} dx \cr
& or \cr
& = \int_0^1 {\frac{x}{{\sqrt {{{\left( 2 \right)}^2} - {{\left( {\sqrt 3 {x^2}} \right)}^2}} }}} dx \cr
& {\text{substitute }}u = \sqrt 3 {x^2},{\text{ }}du = 2\sqrt 3 xdx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 1,{\text{ }}u = \sqrt 3 {\left( 1 \right)^2} = \sqrt 3 \cr
& {\text{if }}x = 0,{\text{ }}u = \sqrt 3 {\left( 0 \right)^2} = 0 \cr
& {\text{so}} \cr
& = \int_0^1 {\frac{x}{{\sqrt {{{\left( 2 \right)}^2} - {{\left( {\sqrt 3 {x^2}} \right)}^2}} }}} dx = \int_0^{\sqrt 3 } {\frac{{\left( {1/2\sqrt 3 } \right)}}{{\sqrt {{{\left( 2 \right)}^2} - {u^2}} }}} du \cr
& {\text{find the antiderivative using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\frac{u}{a} + C,{\text{ }}} \left( {{\text{see page 468}}} \right) \cr
& = \left. {\frac{1}{{2\sqrt 3 }}\left( {si{n^{ - 1}}\frac{u}{2}} \right)} \right|_0^{\sqrt 3 } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{{2\sqrt 3 }}\left( {{{\sin }^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) - {{\sin }^{ - 1}}\left( 0 \right)} \right) \cr
& {\text{simplify}} \cr
& = \frac{1}{{2\sqrt 3 }}\left( {\frac{\pi }{3} - 0} \right) \cr
& = \frac{\pi }{{6\sqrt 3 }} \cr} $$
