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

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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 $z$

$\blacksquare$


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