Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power

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Theorem

$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$

Proof

Aiming for an expression in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

in order to use the technique of Integration by Parts, let:

\(\ds v\)

\(=\)

\(\ds \paren {a x + b}^s\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d v} {\d x}\)

\(=\)

\(\ds a s \paren {a x + b}^{s - 1}\)

Derivative of Power and Derivative of Function of Constant Multiple: Corollary

In order to make $u \dfrac {\d v} {\d x}$ equal to the integrand, let:

\(\ds u\)

\(=\)

\(\ds \frac {\paren {a x + b}^{m - s + 1} } {a s} \paren {p x + q}^n\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d u} {\d x}\)

\(=\)

\(\ds \frac {a \paren {m - s + 1} \paren {a x + b}^{m - s} \paren {p x + q}^n + p n \paren {a x + b}^{m - s + 1} \paren {p x + q}^{n - 1} } {a s}\)

Product Rule for Derivatives and above

\(\ds \)

\(=\)

\(\ds \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a s} \paren {a \paren {m - s + 1} \paren {p x + q} + p n \paren {a x + b} }\)

extracting common factor

\(\ds \)

\(=\)

\(\ds \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a s} \paren {a p x \paren {m - s + 1 + n} + a q \paren {m - s + 1} + p b n}\)

separating out terms in $x$

Select $s$ such that $m - s + n + 1 = 0$, and so $s = m + n + 1$:

\(\ds \frac {\d u} {\d x}\)

\(=\)

\(\ds \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a \paren {m + n + 1} } \paren {a q \paren {m - \paren {m + n + 1} + 1} + p b n}\)

term in $x$ vanishes

\(\ds \)

\(=\)

\(\ds \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a \paren {m + n + 1} } \paren {n \paren {b p - a q} }\)

simplifying

Other instances of $s$ are left as they are, anticipating that they will cancel out later.

Thus:

\(\ds \)

\(\)

\(\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x\)

\(\ds \)

\(=\)

\(\ds \int \frac {\paren {a x + b}^{m - s + 1} } {a s} \paren {p x + q}^n a s \paren {a x + b}^{s - 1} \rd x\)

in the form $\ds \int u \frac {\d v} {\d x} \rd x$

\(\ds \)

\(=\)

\(\ds \frac {\paren {a x + b}^{m - s + 1} } {a s} \paren {p x + q}^n \paren {a x + b}^s\)

in the form $\ds u v - \int v \frac {\d u} {\d x} \rd x$

\(\ds \)

\(\)

\(\, \ds - \, \)

\(\ds \int \paren {a x + b}^s \frac {\paren {a x + b}^{m - s} \paren {p x + q}^{n - 1} } {a \paren {m + n + 1} } \paren {n \paren {p b - a q} } \rd x\)

\(\ds \)

\(=\)

\(\ds \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x\)

Primitive of Constant Multiple of Function

$\blacksquare$

Also defined as

This can also be reported as:

$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$

by interchanging the roles of $m$ and $n$.