1. ## Completing the square

I'm having trouble integrating this: dx/sqrt(5+3x-2x^2). I rearranged it to form dx/sqrt(31/8 - 2(x-3/4)^2) Not sure what to do after that. Should I have not factored the 2 out? Thanks.

2. You have $-2x^2 + 3x +5 = -2(x^2-\frac{3}{2}x-\frac{5}{2})$

$-2(x^2-\frac{3}{2}x+\frac{9}{16}-\frac{9}{16}-\frac{5}{2})$

$-2((x-\frac{3}{4})^2-\frac{49}{16}) = \frac{49}{8} - 2(x-\frac{3}{4})^2$
You can now use a trigo substitution to find the answer.
Keeping the 2 in front would be a nice idea.

3. crap I'm having trouble with the trig substitution. I put (x - 3/4) as u which leaves me with 49/8 - 2(u^2) and then I used u=asin(theta) which gives me with 49/8(1 - 2sin^2(theta)). After this I'm not sure what to do.

4. This leads to $*\int \frac{1}{\sqrt{2}}\frac{dx}{\sqrt{\frac{49}{16}-(x-\frac{3}{4})^2}}$*
You should use $\frac{49}{16}sin(\theta) = x$*
You know these integrals are very tricky and I may make some mistakes.
I'm not too bad at understanding but making no mistakes is not one of my aptitudes.
You can check yourself with the online integrator of mathematica.
It gives $*\frac{1}{\sqrt{2}}arcsin(\frac{4}{7}x-\frac{3}{7}) +C$

5. How did you get from 49/8 to 49/16 and pulling out 1/sqrt(2). And yeah your answer was right.

6. You factor a 2 out front.
Then you can simply get it out of the sqrt and out of the integral.

7. i'm an idiot. thanks bro.