Primitive of x by Root of a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int x \sqrt {a x^2 + b x + c} \rd x = \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac {b \paren {2 a x + b} \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \paren {4 a c - b^2} } {16 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
Proof
\(\ds \) | \(\) | \(\ds \int x \sqrt {a x^2 + b x + c} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \rd x\) | Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a}\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac b {2 a} \paren {\frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } }\) | Primitive of $\sqrt {a x^2 + b x + c}$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt {a x^2 + b x + c} }^3} {3 a}\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac {b \paren {2 a x + b} \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \paren {4 a c - b^2} } {16 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) | simplifying |
$\blacksquare$
Also see
- For $a = 0$, see Primitive of $x \sqrt {a x + b}$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.286$