Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power
Jump to navigation
Jump to search
This article has been identified as a candidate for Featured Proof status. If you do not believe that this proof is worthy of being a Featured Proof, please state your reasons on the talk page. To discuss this page in more detail, feel free to use the talk page. |
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 $n$.