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

Theorem

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

Then for $x \in \R$ such that $a x^2 + b x + c > 0$:

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

Proof

First:

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

Then:

 $\ds \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} }$ $=$ $\ds \frac 1 {2 a} \int \frac {2 a x \rd x} {\sqrt {a x^2 + b x + c} }$ $\ds$ $=$ $\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {\sqrt {a x^2 + b x + c} }$ $\ds$ $=$ $\ds \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\sqrt {a x^2 + b x + c} } - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$ Linear Combination of Integrals $\ds$ $=$ $\ds \frac 1 {2 a} \int \frac {\d z} {\sqrt z} - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$ Integration by Substitution $\ds$ $=$ $\ds \frac 1 {2 a} 2 \sqrt z - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$ Primitive of Power $\ds$ $=$ $\ds \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$ substituting for $z$

$\blacksquare$