Primitive of Root of a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int \sqrt {a x^2 + b x + c} \rd x = \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
Proof
Let:
\(\ds z\) | \(=\) | \(\ds \paren {2 a x + b}^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 4 a \paren {2 a x + b}\) | Derivative of Power and Chain Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 4 a \sqrt z\) |
Suppose $a > 0$.
Then we have:
\(\ds \) | \(\) | \(\ds \int \sqrt {a x^2 + b x + c} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \sqrt {\frac {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } {4 a} } \rd x\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\sqrt {z + \paren {4 a c - b^2} } \rd z} {\paren {2 \sqrt a} \paren {4 a \sqrt z} }\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {8 a \sqrt a} \int \frac {\sqrt {z + \paren {4 a c - b^2} } \rd z} {\sqrt z}\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {8 a \sqrt a} \paren {\sqrt z \sqrt {z + \paren {4 a c - b^2} } + \frac {4 a c - b^2} 2 \int \frac {\d z} {\sqrt z \sqrt {z + \paren {4 a c - b^2} } } }\) | Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt z \sqrt {z + \paren {4 a c - b^2} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {\d z} {\sqrt z \sqrt {z + \paren {4 a c - b^2} } }\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b} \sqrt {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {4 a \rd x} {\sqrt {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } }\) | substituting back for $z$ and $\d z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b} \paren {2 \sqrt a \sqrt {a x^2 + b x + c} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {4 a \rd x} {2 \sqrt a \sqrt {a x^2 + b x + c} }\) | substituting back for $\sqrt {a x^2 + b x + c}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) | simplifying |
When $a < 0$, the above does not work, as we cannot take the square root of a negative number.
Hence in this case:
\(\ds \) | \(\) | \(\ds \int \sqrt {a x^2 + b x + c} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \sqrt {\frac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a} } \rd x\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \sqrt {\frac {\paren {b^2 - 4 a c} - \paren {2 a x + b}^2} {-4 a} } \rd x\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\sqrt {\paren {b^2 - 4 a c} - z} \rd z} {\paren {2 \sqrt {-a} } \paren {4 a \sqrt z} }\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {8 a \sqrt {-a} } \int \frac {\sqrt {-z + \paren {b^2 - 4 a c} } \rd z} {\sqrt z}\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {8 a \sqrt {-a} } \paren {\sqrt z \sqrt {-z + \paren {b^2 - 4 a c} } + \frac {b^2 - 4 a c} 2 \int \frac {\d z} {\sqrt z \sqrt {-z + \paren {b^2 - 4 a c} } } }\) | Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt z \sqrt {\paren {b^2 - 4 a c} - z} } {8 a \sqrt {-a} } + \frac {b^2 - 4 a c} {16 a \sqrt {-a} } \int \frac {\d z} {\sqrt z \sqrt {\paren {b^2 - 4 a c} - z} }\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b} \sqrt {\paren {b^2 - 4 a c} - \paren {2 a x + b}^2} } {8 a \sqrt {-a} } + \frac {b^2 - 4 a c} {16 a \sqrt {-a} } \int \frac {4 a \rd x} {\sqrt {\paren {b^2 - 4 a c} - \paren {2 a x + b}^2} }\) | substituting back for $z$ and $\d z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b} \paren {2 \sqrt {-a} \sqrt {a x^2 + b x + c} } } {8 a \sqrt {-a} } + \frac {b^2 - 4 a c} {16 a \sqrt {-a} } \int \frac {4 a \rd x} {2 \sqrt {-a} \sqrt {a x^2 + b x + c} }\) | substituting back for $\sqrt {a x^2 + b x + c}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {b^2 - 4 a c} {-8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} +\frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) | arranging it into its final form |
$\blacksquare$
Sources
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.37$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.285$