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

Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\displaystyle \int \sqrt {a x^2 + b x + c} \rd x = \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$

Proof

Let:

 $\displaystyle z$ $=$ $\displaystyle \paren {2 a x + b}^2$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d z} {\d x}$ $=$ $\displaystyle 4 a \paren {2 a x + b}$ Derivative of Power and Chain Rule for Derivatives $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d z} {\d x}$ $=$ $\displaystyle 4 a \sqrt z$ Derivative of Power and Chain Rule for Derivatives

Then:

 $\displaystyle$  $\displaystyle \int \sqrt {a x^2 + b x + c} \rd x$ $\displaystyle$ $=$ $\displaystyle \int \sqrt {\frac {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } {4 a} } \rd x$ Completing the Square $\displaystyle$ $=$ $\displaystyle \int \frac {\sqrt {z + \paren {4 a c - b^2} } \rd z} {\paren {2 \sqrt a} \paren {4 a \sqrt z} }$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 {8 a \sqrt a} \int \frac {\sqrt {z + \paren {4 a c - b^2} } \rd z} {\sqrt z}$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac 1 {8 a \sqrt a} \paren {\sqrt z \sqrt {z + \paren {4 a c - b^2} } + \frac {4 a c - b^2} 2 \int \frac {\d z} {\sqrt z \sqrt {z + \paren {4 a c - b^2} } } }$ Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt z \sqrt {z + \paren {4 a c - b^2} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {\d z} {\sqrt z \sqrt {z + \paren {4 a c - b^2} } }$ multiplying out $\displaystyle$ $=$ $\displaystyle \frac {\paren {2 a x + b} \sqrt {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {4 a \rd x} {\sqrt {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } }$ substituting back for $z$ and $\d z$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {2 a x + b} \paren {2 \sqrt a \sqrt {a x^2 + b x + c} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {4 a \rd x} {2 \sqrt a \sqrt {a x^2 + b x + c} }$ substituting back for $\sqrt {a x^2 + b x + c}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$ simplifying

$\blacksquare$