Primitive of p x + q over Root of a x squared plus 2 b x plus c/Proof 1

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Theorem

$\ds \int \dfrac {p x + q} {\sqrt {a x^2 + 2 b x + c} } \rd x = \dfrac p a \sqrt {a x^2 + 2 b x + c} + \paren {q - \dfrac {p b} a} \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }$


Proof

\(\ds \int \dfrac {p x + q} {\sqrt {a x^2 + 2 b x + c} } \rd x\) \(=\) \(\ds \dfrac p {2 a} \int \dfrac {2 a x + \frac {2 a q} p} {\sqrt {a x^2 + 2 b x + c} } \rd x\)
\(\ds \) \(=\) \(\ds \dfrac p {2 a} \int \dfrac {2 a x + 2 b - 2 b + \frac {2 a q} p} {\sqrt {a x^2 + 2 b x + c} } \rd x\)
\(\ds \) \(=\) \(\ds \dfrac p {2 a} \int \dfrac {2 a x + 2 b} {\sqrt {a x^2 + 2 b x + c} } \rd x + \dfrac p {2 a} \paren {\dfrac {2 a q} p - 2 b} \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \dfrac p {2 a} \int \dfrac {2 a x + 2 b} {\sqrt {a x^2 + 2 b x + c} } \rd x + \paren {q - \dfrac {p b} a} \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }\) simplifying
We have from Power Rule for Derivatives that $\map {\dfrac \d {\d x} } {a x^2 + b x + c} = 2 a x + 2 b$, so:
\(\ds \) \(=\) \(\ds \dfrac p {2 a} \paren {2 \sqrt {a x^2 + 2 b x + c} } + \paren {q - \dfrac {p b} a} \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }\) Square Root of Function under Derivative
\(\ds \) \(=\) \(\ds \dfrac p a \sqrt {a x^2 + 2 b x + c} + \paren {q - \dfrac {p b} a} \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }\) simplifying

$\blacksquare$


Sources

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