# Primitive of x over Root of a x squared plus b x plus c

## Theorem

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

Then:

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

## Proof

First:

 $\displaystyle z$ $=$ $\displaystyle a x^2 + b x + c$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d z} {\mathrm d x}$ $=$ $\displaystyle 2 a x + b$ Derivative of Power

Then:

 $\displaystyle \int \frac {x \ \mathrm d x} {\sqrt {a x^2 + b x + c} }$ $=$ $\displaystyle \frac 1 {2 a} \int \frac {2 a x \ \mathrm d x} {\sqrt {a x^2 + b x + c} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 {2 a} \int \frac {\left({2 a x + b - b}\right) \ \mathrm d x} {\sqrt {a x^2 + b x + c} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 {2 a} \int \frac {\left({2 a x + b}\right) \ \mathrm d x} {\sqrt {a x^2 + b x + c} } - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac 1 {2 a} \int \frac {\mathrm d z} {\sqrt z} - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 {2 a} 2 \sqrt z - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$ Primitive of Power $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$ substituting for $z$

$\blacksquare$