I think I am getting closer, but necessarily correct.
integral (10x^5-5x)/square root (x^4-x^2+6) dx
let x = x^4 -x^2 +6
d/dx = 4x^3 - 2x or 2x(2x^2 -1)
1/2 integral x^-1/2 du
1/2 *(u^1/2)/1/2 + C
1/4*u^1/2 + C
1/4 (x^4-x^2+6)^1/2 +C
I think I am getting closer, but necessarily correct.
integral (10x^5-5x)/square root (x^4-x^2+6) dx
let x = x^4 -x^2 +6
d/dx = 4x^3 - 2x or 2x(2x^2 -1)
1/2 integral x^-1/2 du
1/2 *(u^1/2)/1/2 + C
1/4*u^1/2 + C
1/4 (x^4-x^2+6)^1/2 +C
To find $\displaystyle I = \int \frac{10x^3-5x}{\sqrt{x^4-x^2+6}} dx = \frac{5}{2}\int \frac{4x^3-2x}{\sqrt{x^4-x^2+6}} dx$:
Substitute $\displaystyle u = x^4 - x^2 + 6~~~\Rightarrow~~~du = (4x^3 - 2x) dx$.
Then $\displaystyle I = \frac{5}{2} \int \frac{1}{\sqrt{u}} du = \frac{5}{4}\sqrt{u} + C = \frac{5}{4}\sqrt{x^4 - x^2 + 6} + C$.
OK. Right at the beginning. I decided to "bring outside" a factor of $\displaystyle 5/2$ on the first line. Like this:
$\displaystyle \int \frac{10x^3-5x}{\sqrt{x^4-x^2+6}} dx ~~=~~ \frac{5}{2} \cdot \int \frac{(\frac{2}{5})\left(10x^3-5x\right)}{\sqrt{x^4-x^2+6}} dx ~~=~~ \frac{5}{2} \cdot \int \frac{(4x^3-2x)}{\sqrt{x^4-x^2+6}} dx$
Hopefully, from that line above, you can see how I did it. The more important question, I guess, is why I did it!
The reason is, because then it's easy to see the substitution you should use.
After you "bring out" the $\displaystyle 5/2$, the remaining integral has a particular, special form:
$\displaystyle \int \frac{(4x^3-2x)}{\sqrt{x^4-x^2+6}} dx ~~=~~ \int f(u) \left( \frac{du}{dx} \right) dx$
Here, $\displaystyle u = x^4 - x^2 + 6;$
$\displaystyle f(u) = \frac{1}{\sqrt{u}} ~=~ \frac{1}{\sqrt{x^4 - x^2 + 6}};$
$\displaystyle \left( \frac{du}{dx}\right) = (4x^3 - 2x)$.
This whole process of switching (or substituting) $\displaystyle u$ for $\displaystyle x$ is called 'integration by substitution'.
Then, it becomes easy to finish the integral, because you can say:
$\displaystyle \int f(u) \left( \frac{du}{dx} \right) dx ~~=~~ \int f(u) du$
and because you know $\displaystyle f(u) = \frac{1}{\sqrt{u}} = u^{-1/2}$, you can integrate it using the normal rule:
$\displaystyle \int u^n du ~~=~~ \frac{u^{n+1}}{n+1} + C$.
- - -
I hope that helps to explain what I was doing. Still, if there's a specific bit you don't understand, I'll "zoom in" on it and explain further.
$\displaystyle \int {\frac{{10x^3 - 5x}}
{{\sqrt {x^4 - x^2 + 6} }}~dx} = 5\int {\frac{{2x^3 - x}}
{{\sqrt {x^4 - x^2 + 6} }}~dx} = \frac{5}
{2}\int {\frac{{4x^3 - 2x}}
{{\sqrt {x^4 - x^2 + 6} }}~dx}$
It's faster if you only take $\displaystyle u=\sqrt{x^4 - x^2 + 6}$
$\displaystyle I = \frac{5}{2} \int \frac{1}{\sqrt{u}}du$
$\displaystyle I = \frac{5}{2} . \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$
$\displaystyle I = \frac{5}{2} . \frac{\sqrt{u}}{\frac{1}{2}} $
$\displaystyle I = 5 . \sqrt{u} $
$\displaystyle I = 5 . \sqrt{x^4-x^2+6} $