# Primitive of Reciprocal of Root of a x squared plus b x plus c/Positive Discriminant

## Theorem

Let $a \in \R_{\ne 0}$.

Let $b^2 - 4 a c > 0$.

Then:

$\displaystyle \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac 1 {\sqrt a} \map \ln {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C$

## Proof

First:

 $\displaystyle a x^2 + b x + c$ $=$ $\displaystyle \frac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}$ Completing the Square $(1):\quad$ $\displaystyle \implies \ \$ $\displaystyle \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$ $=$ $\displaystyle \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }$

Put:

 $\displaystyle z$ $=$ $\displaystyle 2 a x + b$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d z} {\d x}$ $=$ $\displaystyle 2 a$ Derivative of Power

Let $D = b^2 - 4 a c$.

Thus:

 $\displaystyle \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$ $=$ $\displaystyle \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }$ from $(1)$ $\displaystyle$ $=$ $\displaystyle \int \frac {\d z} {\sqrt a \sqrt {z^2 - D} }$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - D} }$ Primitive of Constant Multiple of Function

Let $b^2 - 4 a c > 0$.

Then:

 $\displaystyle D$ $>$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle D$ $=$ $\displaystyle q^2$ for some $q \in \R$ $\displaystyle \leadsto \ \$ $\displaystyle q$ $=$ $\displaystyle \sqrt {b^2 - 4 a c}$ by definition of $D$

Thus:

 $\displaystyle \int \frac {\d x} {\sqrt{a x^2 + b x + c} }$ $=$ $\displaystyle \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - q^2} }$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt a} \map \ln {z + \sqrt {z^2 - q^2} } + C$ Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt a} \map \ln {2 a x + b + \sqrt {\paren {2 a x + b}^2 - b^2 - 4 a c} } + C$ substituting for $z$ and $q$ $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt a} \map \ln {2 a x + b + \sqrt {4 a \paren {a x^2 + b x + c} } } + C$ Completing the Square $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt a} \map \ln {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C$ rearranging

$\blacksquare$