Inverse Quadratic Root Integral

$\begin{aligned} I & = \int \frac 1{\sqrt{\color{#3D99F6}ax^2+bx+c}}dx & \small \color{#3D99F6} \text{If }f(x) \text{ is a perfect square} \\ & = \int \frac 1{\sqrt{\color{#3D99F6}\left(\sqrt ax+\sqrt c\right)^2}}dx & \small \color{#3D99F6} \implies f(x) = \left(\sqrt ax+\sqrt c\right)^2 \\ & = \int \frac 1{\sqrt ax+\sqrt c}dx \\ & = \int \frac 1{\sqrt a\left(x+\sqrt {\frac ca}\right)}dx \\ & = \boxed{\dfrac 1{\sqrt a}\ln \left(x+\sqrt {\frac ca}\right) + \color{#3D99F6} C} & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \end{aligned}$

Rewrite $f(x)$ in the form:

$a \big( x^2+\frac{b}{a}x+\frac{c}{a} \big)$ $a \bigg( x+\frac{b}{2a} \bigg) ^2+\frac{c}{a}-\frac{b^2}{4a^2}$

Perfect square means $\frac{c}{a}-\frac{b^2}{4a^2} = 0$ or

$b^2 = 4ac$

Therefore we get:

$f(x) = a \bigg( x+\sqrt{\frac{c}{a}} \bigg)$

The integral becomes ( for $x > 0$ ) :

$\int \frac{1}{\sqrt{a \bigg( x+\sqrt{\frac{c}{a}} \bigg)^2}} dx = \int \frac{1}{\sqrt{a} \bigg( x+\sqrt{\frac{c}{a}} \bigg)} dx$

Which clearly evaluates to:

$\int \frac{1}{\sqrt{a} \bigg( x+\sqrt{\frac{c}{a}} \bigg)} dx = \frac{1}{\sqrt{a}} \ln \bigg( x+\sqrt{\frac{c}{a}} \bigg) + const.$