$\int (2x^2+4x-2)^{-\frac{3}{2}} \ dx$
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.