Help me find the following integral by using basic formulas of integral $$\int\frac{\sqrt{4+x^2}+2\sqrt{4-x^2}}{\sqrt{16-x^4}}dx$$
i tried to break this expression into two parts(sum of integrals) the first gives me $\arcsin(x/2)$(by formula) as $16-x^4 = (4-x^2)(4+x^2)$ and square roots of $4+x^2$  are cancelled out(i took square root of 4 out of integral in denominator and got $1-(x/2)^2$ and in the second part i've got $$2\int \dfrac{dx}{(4+x)^{1/2}}$$ and what to do next 

Hint:$$\frac { \sqrt { 4+{ x }^{ 2 } } +2\sqrt { 4-{ x }^{ 2 } }  }{ \sqrt { 16-{ x }^{ 4 } }  } =\frac { 1 }{ \sqrt { 4-{ x }^{ 2 } }  } +\frac { 2 }{ \sqrt { 4+{ x }^{ 2 } }  } $$ then substite $x=2\sin { \theta  } $
