Primitive of p x + q over Root of a x squared plus 2 b x plus c/Proof 2
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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
| \(\ds \int \dfrac {p x + q} {\sqrt {a x^2 + 2 b x + c} } \rd x\) | \(=\) | \(\ds p \int \dfrac x {\sqrt {a x^2 + 2 b x + c} } \rd x + q \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }\) | Linear Combination of Primitives | |||||||||||
| \(\ds \) | \(=\) | \(\ds p \paren {\frac {\sqrt {a x^2 + 2 b x + c} } a - \frac {2 b} {2 a} \int \frac {\d x} {\sqrt {a x^2 + 2 b x + c} } } + q \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }\) | Primitive of $\dfrac x {\sqrt {a x^2 + 2 b x + c} }$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac p a \sqrt {a x^2 + 2 b x + c} + \paren {q - \dfrac {p b} a} \int \dfrac {\d x} {\sqrt {a x^2 + 2 b x + c} }\) | simplifying |
$\blacksquare$