Primitive of Reciprocal of Cube of Root of a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} = \frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + C$
Proof
For $a > 0$:
\(\ds \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) | \(=\) | \(\ds \int \frac {\d x} {\paren {\sqrt {\frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a} } }^3}\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {8 a \sqrt a \rd x} {\paren {\sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} }^3}\) | simplifying |
Then:
\(\ds z\) | \(=\) | \(\ds 2 a x + b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds 2 a\) | Derivative of Power | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {8 a \sqrt a \rd x} {\paren {\sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} }^3}\) | \(=\) | \(\ds \int \frac {8 a \sqrt a \rd z} {2 a \paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) | Primitive of Constant Multiple of Function |
Let $4 a c - b^2 > 0$.
Then:
\(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) | \(=\) | \(\ds 4 \sqrt a \paren {\frac z {\paren {4 a c - b^2} \sqrt {z^2 + 4 a c - b^2} } } + C\) | Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$ |
Let $4 a c - b^2 < 0$.
Then:
\(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) | \(=\) | \(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 - \paren {b^2 - 4 a c} } }^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 \sqrt a \paren {\frac {-z} {\paren {b^2 - 4 a c} \sqrt {z^2 - \paren {b^2 - 4 a c} } } } + C\) | Primitive of $\dfrac 1 {\paren {\sqrt {x^2 - a^2} }^3}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 \sqrt a \paren {\frac z {\paren {4 a c - b^2} \sqrt {z^2 + 4 a c - b^2} } } + C\) | simplifying |
Thus in both cases the same result applies, and so:
\(\ds 4 \sqrt a \int \frac {\d z} {\paren {\sqrt {z^2 + 4 a c - b^2} }^3}\) | \(=\) | \(\ds \frac {4 \sqrt a \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {\paren {2 a x + b}^2 + 4 a c - b^2} } + C\) | substituting for $z$ and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \sqrt a \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {4 a \paren {a x^2 + b x + c} } } + C\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } + C\) | simplifying |
$\Box$
This needs considerable tedious hard slog to complete it. In particular: Similarly for $a < 0$, unless there's a quick way to cover both cases To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x^2 + bx + c}$: $14.290$