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

$\blacksquare$


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Sources