<s>[INST] How to evaluate $\int\frac{1}{3+4x+4x^2}$? 
How to evaluate
  $$\int\frac{1}{3+4x+4x^2}\quad ?
$$

This is what I've done so far:
$$
\frac{1}{4} \int\frac{1}{x^2+x+\frac{3}{4}}
=\frac{1}{4} \int\frac{1}{(x+\frac{1}{2})^2 + \frac{1}{2}}
$$
$$ 
y = \frac{1}{a}\arctan\frac{u}{a},\quad \frac{dy}{du} = \frac{1}{a^2 + u^2}
$$
$$
a = \sqrt{\frac{1}{2}},\quad
u = (x+\frac{1}{2}),\quad \frac{du}{dx} = 1
$$
so
$$ \frac{dy}{dx} = \frac{dy}{du} \cdot 1
$$
I don't know how to proceed.
Could I also have some help with
$$
\int\frac{1}{\sqrt{-4x^2-4x+3}}\quad ?
$$
 [/INST] HINT:Complete the squares and use the results 
$$\int\frac{1}{x^2 + a^2}dx= \frac{1}{a}\arctan \frac{x}{a}+C $$
and
$$\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \frac{x}{a}+C$$.
</s>