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

## Theorem

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

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

Then:

$\displaystyle \int \frac {\d x} {a x^2 + b x + c} = \frac {-2} {2 a x + b} + C$

## Proof

First:

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

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

Then:

 $\displaystyle \int \frac {\d x} {a x^2 + b x + c}$ $=$ $\displaystyle \int \frac {4 a \rd x} {\paren {2 a x + b}^2}$ from $(1)$ $\displaystyle$ $=$ $\displaystyle \frac {-4 a} {2 a \paren {2 a x + b} } + C$ Primitive of $\dfrac 1 {\paren {a x + b}^2}$ $\displaystyle$ $=$ $\displaystyle \dfrac {-2} {2 a x + b} + C$ simplifying

$\blacksquare$