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

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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

$\blacksquare$


Sources