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
$$
\int\frac{1}{\sqrt{-4x^2-4x+3}}\quad ?
$$
 A: What can be done is a double U-Substitution
We have $$\int\frac{1}{3+4x+4x^2}dx$$
By completing the square we see $$\int\frac{1}{3+4x+4x^2}dx = \int\frac{1}{(2x+1)^2+2}dx$$
Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$\frac{1}{2}\int\frac{1}{u^2+2}du = \frac{1}{2}\int\frac{1}{2(\frac{u^2}{2}+1)}du = \frac{1}{4}\int\frac{1}{(\frac{u^2}{2}+1)}du$$
Now we substitute $s = \frac{u}{\sqrt{2}}$ and $ds = \frac{1}{\sqrt{2}}du$
 in the integrand. $$\frac{1}{2\sqrt{2}}\int\frac{1}{s^2+1}ds$$
$$\int\frac{1}{s^2+1}ds = \tan^{-1}(s)$$ So we are left with $$\frac{\tan^{-1}(s)}{2\sqrt{2}}$$
Back substituting $s = \frac{u}{\sqrt{2}}$, $$\frac{\tan^{-1}(s)}{2\sqrt{2}} = \frac{\tan^{-1}(\frac{u}{\sqrt{2}})}{2\sqrt{2}}$$
Now back substitute $u = 2x+1$ and $$\frac{\tan^{-1}(\frac{u}{\sqrt{2}})}{2\sqrt{2}} = \frac{\tan^{-1}(\frac{2x+1}{\sqrt{2}})}{2\sqrt{2}}$$, 
so our final answer is $$\boxed{\frac{\tan^{-1}(\frac{2x+1}{\sqrt{2}})}{2\sqrt{2}}+ C}$$
A: 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
$$\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \frac{x}{a}+C$$.
