nyuuzyou/wikis huggingface dataset:
Primitive of Power of a x + b over Power of p x + q

== Theorem ==
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.112$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.8.$
Category:Primitive of Power of a x + b over Power of p x + q
Category:Primitives involving a x + b and p x + q
Primitive of Power of a x + b over Power of p x + q/Formulation 1

== Theorem ==
:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac {\paren {a x + b}^{m + 1} } {\paren {p x + q}^{n - 1} } + \paren {n - m - 2} a \int \frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } \rd x}$
== Proof ==
From Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - \paren {m + n + 2} a \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$
Setting $n := -n$:
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.112$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.8.$
Category:Primitive of Power of a x + b over Power of p x + q
Primitive of Power of a x + b over Power of p x + q/Formulation 2

== Theorem ==
:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$
== Proof ==
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} a} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$
Setting $n := -n$:
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.112$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.8.$
Category:Primitive of Power of a x + b over Power of p x + q
Primitive of Power of a x + b over Power of p x + q/Formulation 3

== 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:
Then:
Thus:
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.112$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.8.$
Category:Primitive of Power of a x + b over Power of p x + q
Primitive of Power of p x + q by Root of a x + b

== Theorem ==
:$\ds \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt{a x + b} } \rd x$
== Proof ==
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$
Setting $m := \dfrac 1 2$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.116$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.4.$
Category:Primitives involving Root of a x + b and p x + q
Primitive of Power of p x + q over Root of a x + b

== Theorem ==
:$\ds \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$
== Proof ==
From Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^{m + 1} \paren {p x + q}^n} {\paren {m + n + 1} a} - \frac {n \paren {b p - a q} } {\paren {m + n + 1} a} \int \paren {a x + b}^m \paren {p x + q}^{n - 1} \rd x$
Setting $m := -\dfrac 1 2$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.118$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.6.$
Category:Primitives involving Root of a x + b and p x + q
Primitive of Power of x by Arccosecant of x

== Theorem ==
:$\ds \int x^m \arccsc x \rd x = \begin {cases}
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x + \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x - \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\
\end {cases}$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\rd v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
First let $\arccsc x$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
== Also see ==
* Primitive of $x^m \arcsin x$
* Primitive of $x^m \arccos x$
* Primitive of $x^m \arctan x$
* Primitive of $x^m \arccot x$
* Primitive of $x^m \arcsec x$
Category:Primitives involving Inverse Cosecant Function
Primitive of Power of x by Arccosecant of x over a

== Theorem ==
:$\ds \int x^m \arccsc \frac x a \rd x = \begin {cases}
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a + \dfrac a {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \\
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a - \dfrac a {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0
\end {cases}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.508$
* : $\S 17$: Tables of Special Indefinite Integrals: $(24)$ Integrals Involving Inverse Trigonometric Functions: $17.24.38.$
Category:Primitive of Power of x by Arccosecant of x over a
Category:Primitives involving Inverse Cosecant Function
Primitive of Power of x by Arccosecant of x over a/Proof 1

== Theorem ==
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
Category:Primitive of Power of x by Arccosecant of x over a
Primitive of Power of x by Arccosecant of x over a/Proof 2

== Theorem ==
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\rd v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2} 0$.
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
Category:Primitive of Power of x by Arccosecant of x over a
Primitive of Power of x by Arccosine of a x

== Theorem ==
:$\ds \int x^n \arccos a x \rd x = \frac {x^{n + 1} } {n + 1} \arccos a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} }$
for $n \ne -1$.
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^n \arcsin a x$
* Primitive of $x^n \arctan a x$
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $100$.
Category:Primitives involving Inverse Cosine Function
Primitive of Power of x by Arccosine of x

== Theorem ==
:$\ds \int x^m \arccos x \rd x = \frac {x^{m + 1} } {m + 1} \arccos x + \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {1 - x^2} }$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arcsin x$
* Primitive of $x^m \arctan x$
* Primitive of $x^m \arccot x$
* Primitive of $x^m \arcsec x$
* Primitive of $x^m \arccsc x$
Category:Primitives involving Inverse Cosine Function
Primitive of Power of x by Arccosine of x over a

== Theorem ==
:$\ds \int x^m \arccos \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arccos \frac x a + \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.504$
Category:Primitive of Power of x by Arccosine of x over a
Category:Primitives involving Inverse Cosine Function
Primitive of Power of x by Arccosine of x over a/Proof 1

== Theorem ==
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arccosine of x over a
Primitive of Power of x by Arccosine of x over a/Proof 2

== Theorem ==
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitives involving Inverse Cosine Function
Primitive of Power of x by Arccotangent of x

== Theorem ==
:$\ds \int x^m \arccot x \rd x = \frac {x^{m + 1} } {m + 1} \arccot x + \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + 1}$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arcsin x$
* Primitive of $x^m \arccos x$
* Primitive of $x^m \arctan x$
* Primitive of $x^m \arcsec x$
* Primitive of $x^m \arccsc x$
Category:Primitives involving Inverse Cotangent Function
Primitive of Power of x by Arccotangent of x over a

== Theorem ==
:$\ds \int x^m \arccot \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arccot \frac x a + \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.506$
Category:Primitive of Power of x by Arccotangent of x over a
Category:Primitives involving Inverse Cotangent Function
Primitive of Power of x by Arccotangent of x over a/Proof 1

== Theorem ==
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arccotangent of x over a
Primitive of Power of x by Arccotangent of x over a/Proof 2

== Theorem ==
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arccotangent of x over a
Primitive of Power of x by Arcsecant of x

== Theorem ==
:$\ds \int x^m \arcsec x \rd x = \begin {cases}
\dfrac {x^{m + 1} } {m + 1} \arcsec x - \dfrac 1 {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arcsec x < \dfrac \pi 2 \\
\dfrac {x^{m + 1} } {m + 1} \arcsec x + \dfrac 1 {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : \dfrac \pi 2 < \arcsec x < \pi \\
\end {cases}$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
First let $\arcsec x$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
Similarly, let $\arcsec x$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
== Also see ==
* Primitive of $x^m \arcsin x$
* Primitive of $x^m \arccos x$
* Primitive of $x^m \arctan x$
* Primitive of $x^m \arccot x$
* Primitive of $x^m \arccsc x$
Category:Primitives involving Inverse Secant Function
Primitive of Power of x by Arcsecant of x over a

== Theorem ==
: $\ds \int x^m \arcsec \frac x a \rd x = \begin {cases}
\dfrac {x^{m + 1} } {m + 1} \arcsec \dfrac x a - \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {x^{m + 1} } {m + 1} \arcsec \dfrac x a + \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end {cases}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.507$
Category:Primitive of Power of x by Arcsecant of x over a
Category:Primitives involving Inverse Secant Function
Primitive of Power of x by Arcsecant of x over a/Proof 1

== Theorem ==
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arcsecant of x over a
Primitive of Power of x by Arcsecant of x over a/Proof 2

== Theorem ==
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arcsecant of x over a
Primitive of Power of x by Arcsine of a x

== Theorem ==
:$\ds \int x^n \arcsin a x \rd x = \frac {x^{n + 1} } {n + 1} \arcsin a x - \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} }$
for $n \ne -1$.
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^n \arccos a x$
* Primitive of $x^n \arctan a x$
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $99$.
Category:Primitives involving Inverse Sine Function
Primitive of Power of x by Arcsine of x

== Theorem ==
:$\ds \int x^m \arcsin x \rd x = \frac {x^{m + 1} } {m + 1} \arcsin x - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {1 - x^2} }$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arccos x$
* Primitive of $x^m \arctan x$
* Primitive of $x^m \arccot x$
* Primitive of $x^m \arcsec x$
* Primitive of $x^m \arccsc x$
Category:Primitives involving Inverse Sine Function
Primitive of Power of x by Arcsine of x over a

== Theorem ==
:$\ds \int x^m \arcsin \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arcsin \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {\sqrt {a^2 - x^2} }$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.503$
Category:Primitive of Power of x by Arcsine of x over a
Category:Primitives involving Inverse Sine Function
Primitive of Power of x by Arcsine of x over a/Proof 1

== Theorem ==
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arcsine of x over a
Primitive of Power of x by Arcsine of x over a/Proof 2

== Theorem ==
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arctan \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arcsine of x over a
Primitive of Power of x by Arctangent of a x

== Theorem ==
:$\ds \int x^n \arctan a x \rd x = \frac {x^{n + 1} } {n + 1} \arctan a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {a^2 x^2 + 1}$
for $n \ne -1$.
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^n \arcsin a x$
* Primitive of $x^n \arccos a x$
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $101$.
Category:Primitives involving Inverse Tangent Function
Primitive of Power of x by Arctangent of x

== Theorem ==
:$\ds \int x^m \arctan x \rd x = \frac {x^{m + 1} } {m + 1} \arctan x - \frac 1 {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + 1}$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arcsin x$
* Primitive of $x^m \arccos x$
* Primitive of $x^m \arccot x$
* Primitive of $x^m \arcsec x$
* Primitive of $x^m \arccsc x$
Category:Primitives involving Inverse Tangent Function
Primitive of Power of x by Arctangent of x over a

== Theorem ==
:$\ds \int x^m \arctan \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arctan \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.505$
Category:Primitive of Power of x by Arctangent of x over a
Category:Primitives involving Inverse Tangent Function
Primitive of Power of x by Arctangent of x over a/Proof 1

== Theorem ==
== Proof ==
Recall:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitive of Power of x by Arctangent of x over a
Primitive of Power of x by Arctangent of x over a/Proof 2

== Theorem ==
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arcsin \dfrac x a$
* Primitive of $x^m \arccos \dfrac x a$
* Primitive of $x^m \arccot \dfrac x a$
* Primitive of $x^m \arcsec \dfrac x a$
* Primitive of $x^m \arccsc \dfrac x a$
Category:Primitives involving Inverse Tangent Function
Primitive of Power of x by Cosine of a x

== Theorem ==
:$\ds \int x^m \cos a x \rd x = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \int x^{m - 2} \cos a x \rd x$
=== Corollary ===
== Proof ==
=== Lemma ===
From :
:$(1): \quad \ds \int x^{m - 1} \sin a x \rd x = \frac {-x^{m - 1} \cos a x} a + \frac {m - 1} a \int x^{m - 2} \cos a x \rd x$
So:
== Also see ==
* Primitive of $x^m \sin a x$
== Sources ==
* : $\S 14$: Integrals involving $\cos a x$: $14.394$
Category:Primitive of Power of x by Cosine of a x
Category:Primitives involving Cosine Function
Primitive of Power of x by Cosine of a x/Corollary

== Theorem ==
:$\ds \int x^m \cos a x \rd x = \sum_{k \mathop = 1}^{m + 1} \paren {m^{\underline {k - 1} } \frac {x^{m + 1 - k} } {a^k} \map {\sin} {x + \dfrac {\pi} 2 \paren {k - 1} } }$
where $m^{\underline {k - 1} }$ denotes the $k - 1$th falling factorial of $m$.
== Proof ==
Category:Primitives involving Cosine Function
Primitive of Power of x by Cosine of a x/Lemma

== Theorem ==
:$\ds \int x^m \cos a x \rd x = \frac {x^m \sin a x} a - \frac m a \int x^{m - 1} \sin a x \rd x$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \sin a x$
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $81$.
Category:Primitive of Power of x by Cosine of a x
Category:Primitives involving Cosine Function
Primitive of Power of x by Exponential of a x

== Theorem ==
Let $n$ be a positive integer.
Let $a$ be a non-zero real number.
Then:
where $n^{\underline k}$ denotes the $k$th falling factorial power of $n$.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$
$\map P 0$ is true, as from Primitive of $e^{a x}$:
:$\ds \int e^{a x} \rd x = \frac {e^{a x} } a$
=== Basis for the Induction ===
$\map P 1$ is the case:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
So this is our induction hypothesis:
:$\ds \int x^r e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^r \paren {\paren {-1}^k \frac {r^{\underline k} x^{r - k} } {a^k} } + C$
Then we need to show:
:$\ds \int x^{r + 1} e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^{r + 1} \paren {\paren {-1}^k \frac {\paren {r + 1}^{\underline k} x^{r + 1 - k} } {a^k} } + C$
=== Induction Step ===
This is our induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \int x^n e^{a x} \rd x = \frac {e^{a x} } a \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} x^{n - k} } {a^k} } + C$
== Examples ==
== Also see ==
* Primitive of $\dfrac {e^{a x} } x$ for the case where $n = -1$
* Primitive of $\dfrac {e^{a x} } {x^n}$
== Sources ==
* : $\S 14$: Integrals involving $e^{a x}$: $14.512$
Category:Primitive of Power of x by Exponential of a x
Category:Primitives involving Exponential Function
Primitive of Power of x by Exponential of a x/Examples/x cubed by e^-x

== Example of Use of Primitive of $x^n e^{a x}$ ==
:$\ds \int x^3 e^{-x} \rd x = -e^{-x} \paren {x^3 + 3 x^2 + 6 x + 6} + C$
== Proof ==
From Primitive of $x^n e^{a x}$:
Hence:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $26$.
Category:Examples of Use of Primitive of Power of x by Exponential of a x