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
OK. Right at the beginning. I decided to "bring outside" a factor of on the first line. Like this:
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 , the remaining integral has a particular, special form:
Here,
.
This whole process of switching (or substituting) for is called 'integration by substitution'.
Then, it becomes easy to finish the integral, because you can say:
and because you know , you can integrate it using the normal rule:
.
- - -
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.