Solve $\displaystyle \int {\frac{{dx}}
{{\sqrt {ax^2 + bx + c} }}} .\quad b,c \in \mathbb{R},\,a > 0.$
WLOG I'll consider $\displaystyle \displaystyle \int \frac{1}{\sqrt{x^2 + Bx + C}} \, dx$.
Then I reckon the solution what you was lookin' for uses the substitution $\displaystyle \sqrt{x^2 + Bx + C} = x + z$.
$\displaystyle \ln (B + 2x + 2\sqrt{x^2 + Bx + C})$ falls out easy peasy.
Following from my previous post (and probably totally not necessary for Krizalid):
WLOG let a = 1, b = B and c = C. You get $\displaystyle \displaystyle u = 2x + B + 2\sqrt {x^2 + Bx + C}$. The substitution I used (the standard Euler substitution, just to set the record straight) is $\displaystyle \displaystyle u + x = \sqrt{x^2 + Bx + C}$. Arguably simpler .....?? Arguably easier ....?? But ...... intuitively way less obvious .. ..??
Personally, I prefer $\displaystyle \displaystyle u + x = \sqrt {x^2 + Bx + C}$ ..... easy to remember for this type of integrand viz. $\displaystyle f(\sqrt{x^2 + Bx + C})$ and the algebra for getting the stuff that's needed (dx and x in terms of u) is easy.
Eg. To solve for x in terms of u: square both sides of $\displaystyle \displaystyle \sqrt{x^2 + Bx + C} = x + u$, subtract $\displaystyle x^2$ from both sides, group x-terms, solve for x.
Note: In my original post I used z (not u) as my new variable, but u's OK I guess.