Primitive of Power of a x + b over Power of p x + q/Formulation 3
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Theorem
- $\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^{n - 1}} \rd x}$
Proof
Let:
\(\ds u\) | \(=\) | \(\ds \paren {a x + b}^m\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds m a \paren {a x + b}^m\) | Power Rule for Derivatives and Derivatives of Function of $a x + b$ |
Then:
\(\ds v\) | \(=\) | \(\ds \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \paren {p x + q}^n\) | Power Rule for Derivatives and Derivatives of Function of $a x + b$ |
Thus:
\(\ds \) | \(\) | \(\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a x + b}^m \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } - \int \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } m a \paren {a x + b}^m \rd x\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m-1} } {\paren {p x + q}^{n - 1} } \rd x}\) | simplification |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.112$