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

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Theorem

$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$


Proof

Aiming for an expression in the form:

$\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$

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

\(\displaystyle v\) \(=\) \(\displaystyle \left({a x + b}\right)^s\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle a s \left({a x + b}\right)^{s-1}\) Derivative of Power and Derivative of Function of Constant Multiple: Corollary


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

\(\displaystyle u\) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle \frac {a \left({m - s + 1}\right) \left({a x + b}\right)^{m - s} \left({p x + q}\right)^n + p n \left({a x + b}\right)^{m - s + 1} \left({p x + q}\right)^{n-1} } {a s}\) Product Rule for Derivatives and above
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a s} \left({a \left({m - s + 1}\right) \left({p x + q}\right) + p n \left({a x + b}\right) }\right)\) extracting common factor
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a s} \left({a p x \left({m - s + 1 + n}\right) + a q \left({m - s + 1}\right) + p b n}\right)\) separating out terms in $x$


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

\(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({a q \left({m - \left({m + n + 1}\right) + 1}\right) + p b n}\right)\) term in $x$ vanishes
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({n \left({p b - a q}\right)}\right)\) simplifying

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


Thus:

\(\displaystyle \) \(\) \(\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \int \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n a s \left({a x + b}\right)^{s-1} \ \mathrm d x\) in the form $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n \left({a x + b}\right)^s\) in the form $\displaystyle u v - \int v \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \int \left({a x + b}\right)^s \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({n \left({p b - a q}\right)}\right) \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x\) Primitive of Constant Multiple of Function

$\blacksquare$


Also defined as

This can also be reported as:

$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^m \left({p x + q}\right)^{n+1}} {\left({m + n + 1}\right) p} + \frac {m \left({b p - a q}\right)} {\left({m + n + 1}\right) p} \int \left({a x + b}\right)^{m-1} \left({p x + q}\right)^n \ \mathrm d x$

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