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