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:

$\displaystyle \int x \sqrt {a x^2 + b x + c} \ \mathrm d x = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac {b \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \left({4 a c - b^2}\right)} {16 a^2} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$


Proof

\(\displaystyle \) \(\) \(\displaystyle \int x \sqrt {a x^2 + b x + c} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x\) Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a}\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac b {2 a} \left({\frac {\left({2 a x + b}\right) \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} } }\right)\) Primitive of $\sqrt {a x^2 + b x + c}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a}\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \frac {b \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \left({4 a c - b^2}\right)} {16 a^2} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }\) simplifying

$\blacksquare$


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