Suppose I have to integrate arbitrary quadratic polynomial in square root
$$\int{\sqrt{ax^2+bx+c}}\,dx$$
And then I use $u$ substitution as this
$$u = \sqrt{-ax^2-bx-c}$$
$$ax^2+bx+(c+u^2) = 0$$
$$x = \frac{-b-\sqrt{b^2 - 4a(c+u^2)}}{2a}$$
Which means it should cancel everything out into
$$\int{\sqrt{u^2}}\,d \frac{-b-\sqrt{b^2 - 4a(c-u^2)}}{2a}$$
$$\int{u}\,d \frac{-b-\sqrt{b^2 - 4a(c-u^2)}}{2a}$$
$$\left[\frac{u^2}{2}\right]_\frac{-b-\sqrt{b^2 - 4a(c-u_0^2)}}{2a}^\frac{-b-\sqrt{b^2 - 4a(c-u_1^2)}}{2a}$$
And if this is correct, I think actually we might be able to derive this formula for both 3rd and 4th-degree polynomial? (because they also have the solution in radical)
It seems like I did something wrong here. Could anyone educate me where did I go wrong?
OK I have seen why it wrong now. The formula I use is incorrect. Thank you