Primitive of Reciprocal of x squared by Cube of Root of a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int \frac {\d x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} = -\frac {a x^2 + 2 b x + c} {c^2 x \sqrt {a x^2 + b x + c} } + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac {3 b} {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$
Proof
\(\ds \) | \(\) | \(\ds \int \frac {\d x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {c \rd x} {c x^2 \paren {\sqrt {a x^2 + b x + c} }^3}\) | multiplying top and bottom by $c$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x^2 \paren {\sqrt {a x^2 + b x + c} }^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} - \frac a c \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {x \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 c \int \frac {\d x} {x^2 \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 c \paren {-\frac {\sqrt {a x^2 + b x + c} } {c x} - \frac b {2 c} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } } - \frac a c \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) | Primitive of $\dfrac 1 {x^2 \sqrt {a x^2 + b x + c} }$ | |||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac b c \paren {\frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} }\) | Primitive of $\dfrac 1 {x \paren {\sqrt {a x^2 + b x + c} }^3}$ | ||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {a x^2 + b x + c} {c^2 x \sqrt {a x^2 + b x + c} } - \frac b {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac {2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) | arranging common denominators for like terms | |||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac {b x} {c^2 x \sqrt {a x^2 + b x + c} } - \frac b {c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } + \frac {b^2} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {a x^2 + 2 b x + c} {c^2 x \sqrt {a x^2 + b x + c} } + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac {3 b} {2 c^2} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }\) | assembling final form |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.294$