Primitive of p x + q over Root of a x + b
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Theorem
- $\ds \int \frac {p x + q} {\sqrt {a x + b} } \rd x = \frac {2 \paren {a p x + 3 a q - 2 b p} } {3 a^2} \sqrt{a x + b}$
Proof
Let:
\(\ds u\) | \(=\) | \(\ds \sqrt{a x + b}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac {u^2 - b} a\) |
Then:
\(\ds \int \frac {p x + q} {\sqrt {a x + b} } \rd x\) | \(=\) | \(\ds \int \paren {p \paren {\frac {u^2 - b} a} + q} \frac {2 u} {a u} \rd x\) | Primitive of Function of $\sqrt {a x + b}$ | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \int \paren {p u^2 - b p + a q} \rd x\) | Linear Combination of Primitives | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \paren {\frac {p u^3} 3 - b p z + a q z} + C\) | Primitive of Power | ||||||||||||
Note that at this point we can assume $u > 0$, otherwise $\sqrt {a x + b}$ would not have been defined. | |||||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \paren {\frac {p \paren {\sqrt {a x + b} }^3} 3 - b p \sqrt {a x + b} + a q \sqrt {a x + b} } + C\) | substituting for $u$ | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {a p x - 2 b p + 3 a q} } {3 a^2} \sqrt {a x + b} + C\) | simplification |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.113$