Primitive of x by Root of a x squared plus b x plus c

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

$\blacksquare$