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:

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


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 $z$

$\blacksquare$


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