Help solving another integral $\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$ $$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$$
Complete the square?
$$\int \frac{1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$$
Not sure what do do next, or if I should try something else?
Big help if you can show as "step-by-step" possible as you can.
Thanks in advance!

@Chandru1:
I'm confused at what happens to the 2 that gets factored out when completing the square (the 2 in the denominator) because it looks like Arturo left it out and instead substituted: 
$\int \frac {1}{(2(x+1)^2-4)^\frac{3}{2}} \ dx$ 
$u = x+1$
 $du =dx$ 
which changes the integral to  $\int \frac {1}{(2(u^2-4))^\frac{3}{2}}$ 
Would I be best factoring it out as a 1/2 in front of the integral  or leaving it in and then using trig. substitution like this?  
$u= \sqrt{2}sec(t)$
 $du = \sqrt{2}sec(t)*tan(t) dt$ 
Which changes it to:
$\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(2sec^2(t)-4))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2(tan^2(t)))^\frac{3}{2}} \ dt$ = $\int \frac {\sqrt{2}sec(t)*tan(t)}{(2tan^3(t))} \ dt$
My $\sqrt{2}$ factors out like yours, but I have no 1/8 so figure I factor it out as a 1/2 earlier like I thought?  But then the constant in front of my integral would be $\frac{\sqrt{2}}{2}$  and you have $\frac{\sqrt{2}}{8}$
I know what's wrong, just not how I went wrong.
 A: Hint. $(2x^{2}+4x-2) = 2(x^{2}+2x +1) -4 = 2(x+1)^{2}-4$. Now subsitute $x+1 = \sqrt{2} \sec{t}$. Then you have $dx = \sqrt{2} \sec{t} \cdot \tan{t} \ dt$. Now, $$2 \cdot \Bigl[(x+1)^{2} -2 \Bigr]= 2 \cdot \Bigl[ 2 \cdot (\sec^{2}{t}-1)\Bigr] = 4 \cdot (\sec^{2}{t}-1)$$.
Now what will you get if you power them by $\frac{3}{2}$. You will get $4^{3/2} \cdot \Bigl[ \tan^{2}{t}\Bigr]^{3/2} = 8 \cdot \tan^{3}{t}$.
And try to evaluate the integral. Simplyfying you should get 
\begin{align*}
\int \frac{1}{(2x^{2}+4x-2)^{-\frac{3}{2}}} \ \textrm{dx} &= \frac{\sqrt{2}}{8} \int\frac{\sec{t} \cdot \tan{t}}{\tan^{3}{t}} \ \textrm{dt} \\ &= \frac{\sqrt{2}}{8} \int\frac{\cos{t}}{\sin^{2}{t}} \ \textrm{dt}
\end{align*}
A: First, factor out a $2$:
$$(2x^2 + 4x - 2) = 2(x^2+2x-1)\quad\text{so}\quad (2x^2+4x-2)^{-3/2} = 2^{-3/2}(x^2+2x-1)^{-3/2}.$$
Do this because you can pull it out of the integral and it will make things easier.
So, up to a constant, this is the same as doing the integral
$$\frac{1}{2^{3/2}}\int\frac{dx}{(x^2+2x-1)^{3/2}}.$$
Completing the square is a good first step. $x^2+2x-1 = (x^2+2x+1) - 2 = (x+1)^2 - 2$. So we get the integral
$$\frac{1}{2^{3/2}}\int\frac{dx}{\bigl( (x+1)^2 - 2\bigr)^{3/2}}.$$
Doing a change of variable $u=x+1$ changes it to
$$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}}.$$
Now try a trigonometric substitution to get rid of that pesky square root in the exponent. Use $\tan^2\theta + 1 = \sec^2\theta$, or $\sec^2\theta - 1 =\tan^2\theta$. So set $u = \sqrt{2}\sec\theta$ to get
$$u^2-2 = 2\sec^2\theta - 2 = 2(\sec^2\theta - 1) = 2\tan^2\theta$$
and $du = \sqrt{2}\sec\theta\tan\theta\,d\theta$, so
$$\frac{1}{2^{3/2}}\int\frac{du}{(u^2-2)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{(2\tan^2\theta)^{3/2}} = \frac{1}{2^{3/2}}\int\frac{\sqrt{2}\sec\theta\tan\theta\,d\theta}{2^{3/2}\tan^3\theta}.$$
Now work the trigonometric integral.
A: Can you see that the expression in the parentheses can be written as $((x+1)^2-2)$. Do we see a difference of squares here?
Next a u sub is required.
$U=(x+1)-\sqrt 2$
Fill in the details.
A partial fraction decomp of $1/((u+2\sqrt 2)u^2)$ will be needed.
And the solution has become as sweet as apple pie!
