nyuuzyou/wikis huggingface dataset:
Primitive of Power of x by Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int x^m \cosh a x \rd x = \frac {x^m \sinh a x} a - \frac m a \int x^{m - 1} \sinh a x \rd x + C$
== 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 $x^m \sinh a x$
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.585$
* : Back endpapers: A Brief Table of Integrals: $122$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(28)$ Integrals Involving $\cosh a x$: $17.28.12.$
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Power of x by Hyperbolic Sine of a x

== Theorem ==
:$\ds \int x^m \sinh a x \rd x = \frac {x^m \cosh a x} a - \frac m a \int x^{m - 1} \cosh a x \rd x$
== 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 $x^m \cosh a x$
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x$: $14.557$
* : Back endpapers: A Brief Table of Integrals: $121$.
Category:Primitives involving Hyperbolic Sine Function
Primitive of Power of x by Inverse Hyperbolic Cosecant of x over a

== Theorem ==
:$\ds \int x^m \arcsch \frac x a \rd x = \begin{cases}
\ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a + \frac a {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x > 0 \\ \\
\ds \frac {x^{m + 1} } {m + 1} \arcsch \frac x a - \frac a {m + 1} \int \frac {x^m} {\sqrt {x^2 + a^2} } \rd x + C & : x < 0
\end{cases}$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \arsinh \dfrac x a$
* Primitive of $x^m \arcosh \dfrac x a$
* Primitive of $x^m \artanh \dfrac x a$
* Primitive of $x^m \arcoth \dfrac x a$
* Primitive of $x^m \arsech \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.677$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.20.$
Category:Primitives involving Inverse Hyperbolic Cosecant Function
Primitive of Power of x by Inverse Hyperbolic Cosine of x over a

== Theorem ==
:$\ds \int x^m \arcosh \frac x a \rd x = \dfrac {x^{m + 1} } {m + 1} \arcosh \dfrac x a - \dfrac 1 {m + 1} \int \dfrac {x^{m + 1} } {\sqrt {x^2 - a^2} } \rd x + C$
where $\arcosh$ denotes the real area hyperbolic cosine.
=== Corollary ===
== 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 $x^m \arsinh \dfrac x a$
* Primitive of $x^m \artanh \dfrac x a$
* Primitive of $x^m \arcoth \dfrac x a$
* Primitive of $x^m \arsech \dfrac x a$
* Primitive of $x^m \arcsch \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.673$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.16.$
Category:Primitive of Power of x by Inverse Hyperbolic Cosine of x over a
Category:Primitives involving Inverse Hyperbolic Cosine Function
Primitive of Power of x by Inverse Hyperbolic Cosine of x over a/Corollary

== Theorem ==
:$\ds \int x^m \paren {-\cosh^{-1} \frac x a} \rd x = \dfrac {x^{m + 1} } {m + 1} \paren {-\cosh^{-1} \frac x a} + \dfrac 1 {m + 1} \int \dfrac {x^{m + 1} } {\sqrt {x^2 - a^2} } \rd x + C$
where $-\cosh^{-1}$ denotes the negative branch of the real inverse hyperbolic cosine multifunction.
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.673$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.16.$
Category:Primitive of Power of x by Inverse Hyperbolic Cosine of x over a
Primitive of Power of x by Inverse Hyperbolic Cotangent of x over a

== Theorem ==
:$\ds \int x^m \arcoth \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \arcoth \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} } {a^2 - x^2} \rd x + C$
== 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 $x^m \arsinh \dfrac x a$
* Primitive of $x^m \arcosh \dfrac x a$
* Primitive of $x^m \artanh \dfrac x a$
* Primitive of $x^m \arsech \dfrac x a$
* Primitive of $x^m \arcsch \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.675$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.18.$
Category:Primitives involving Inverse Hyperbolic Cotangent Function
Primitive of Power of x by Inverse Hyperbolic Secant of x over a

== Theorem ==
:$\ds \int x^m \arsech \frac x a \rd x = \dfrac {x^{m + 1} } {m + 1} \arsech \dfrac x a + \dfrac a {m + 1} \int \dfrac {x^m} {\sqrt {a^2 - x^2} } \rd x + C$
where $\arsech$ denotes the real area hyperbolic secant.
=== Corollary ===
== 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 $x^m \arsinh \dfrac x a$
* Primitive of $x^m \arcosh \dfrac x a$
* Primitive of $x^m \artanh \dfrac x a$
* Primitive of $x^m \arcoth \dfrac x a$
* Primitive of $x^m \arcsch \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.676$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.19.$
Category:Primitive of Power of x by Inverse Hyperbolic Secant of x over a
Category:Primitives involving Inverse Hyperbolic Secant Function
Primitive of Power of x by Inverse Hyperbolic Secant of x over a/Corollary

== Theorem ==
:$\ds \int x^m \paren {-\sech^{-1} \frac x a} \rd x = \dfrac {x^{m + 1} } {m + 1} \paren {-\sech^{-1} \frac x a} - \dfrac a {m + 1} \int \dfrac {x^m} {\sqrt {a^2 - x^2} } \rd x + C$
where $-\sech^{-1}$ denotes the negative branch of the real inverse hyperbolic secant multifunction.
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.676$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.19.$
Category:Primitive of Power of x by Inverse Hyperbolic Secant of x over a
Primitive of Power of x by Inverse Hyperbolic Sine of x over a

== Theorem ==
:$\ds \int x^m \sinh^{-1} \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \sinh^{-1} \frac x a - \frac 1 {m + 1} \int \frac {x^{m + 1} } {\sqrt {x^2 + a^2} } \rd x + C$
== 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 $x^m \cosh^{-1} \dfrac x a$
* Primitive of $x^m \tanh^{-1} \dfrac x a$
* Primitive of $x^m \coth^{-1} \dfrac x a$
* Primitive of $x^m \sech^{-1} \dfrac x a$
* Primitive of $x^m \csch^{-1} \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.672$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.15.$
Category:Primitives involving Inverse Hyperbolic Sine Function
Primitive of Power of x by Inverse Hyperbolic Tangent of x over a

== Theorem ==
:$\ds \int x^m \artanh \frac x a \rd x = \frac {x^{m + 1} } {m + 1} \artanh \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} } {a^2 - x^2} \rd x + C$
== 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 $x^m \arsinh \dfrac x a$
* Primitive of $x^m \arcosh \dfrac x a$
* Primitive of $x^m \arcoth \dfrac x a$
* Primitive of $x^m \arsech \dfrac x a$
* Primitive of $x^m \arcsch \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.674$
* : $\S 17$: Tables of Special Indefinite Integrals: $(34)$ Integrals Involving Inverse Hyperbolic Functions: $17.34.17.$
Category:Primitives involving Inverse Hyperbolic Tangent Function
Primitive of Power of x by Logarithm of a x

== Theorem ==
:$\ds \int x^n \ln a x \rd x = \dfrac {x^{n + 1} } {n + 1} \ln a x - \dfrac {x^{n + 1} } {\paren {n + 2}^2} + C$
for $n \ne -1$.
== 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 $\dfrac {\ln a x} x$ for $n = -1$
* Primitive of $x^n \ln x$
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $110$.
Category:Primitives involving Logarithm Function
Primitive of Power of x by Logarithm of x

== Theorem ==
:$\ds \int x^m \ln x \rd x = \frac {x^{m + 1} } {m + 1} \paren {\ln x - \frac 1 {m + 1} } + C$
where $m \ne -1$.
== 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:
== Example ==
== Also see ==
* Primitive of $\dfrac {\ln x} x$ for the case where $m = -1$.
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.527$
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitives involving Logarithm Function
Primitive of Power of x by Logarithm of x/Example

== Example of Primitive of Power of x by Logarithm of x ==
:$\ds \int_1^n x^m \ln x \rd x = \frac {n^{m + 1} } {m + 1} \paren {\ln n - \frac 1 {m + 1} } + \frac 1 {\paren {m + 1}^2}$
where $m \ne -1$.
== Proof ==
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers
Category:Primitives involving Logarithm Function
Primitive of Power of x by Logarithm of x squared minus a squared

== Theorem ==
:$\ds \int x^m \, \map \ln {x^2 - a^2} \rd x = \frac {x^{m + 1} \map \ln {x^2 - a^2} } {m + 1} - \frac 2 {m + 1} \int \frac {x^{m + 2} } {x^2 - a^2} \rd x$
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$
let:
and let:
Then:
== Also see ==
* Primitive of $x^m \map \ln {x^2 + a^2}$
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.539$
* : $\S 17$: Tables of Special Indefinite Integrals: $(26)$ Integrals Involving $\ln x$: $17.26.15.$
Category:Primitives involving Logarithm Function
Category:Primitives involving x squared minus a squared
Primitive of Power of x by Logarithm of x squared plus a squared

== Theorem ==
:$\ds \int x^m \map \ln {x^2 + a^2} \rd x = \frac {x^{m + 1} \map \ln {x^2 + a^2} } {m + 1} - \frac 2 {m + 1} \int \frac {x^{m + 2} } {x^2 + a^2} \rd x$
== 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 $x^m \map \ln {x^2 - a^2}$
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.539$
* : $\S 17$: Tables of Special Indefinite Integrals: $(26)$ Integrals Involving $\ln x$: $17.26.15.$
Category:Primitives involving Logarithm Function
Category:Primitives involving x squared plus a squared
Primitive of Power of x by Power of Logarithm of x

== Theorem ==
:$\ds \int x^m \ln^n x \rd x = \frac {x^{m + 1} \ln^n x} {m + 1} - \frac n {m + 1} \int x^m \ln^{n - 1} x \rd x + C$
where $m \ne -1$.
== 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 $\dfrac {\ln^n x} x$ for the case where $m = -1$.
* Definite Integral from 0 to 1 of Power of x by Power of Logarithm of x
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.536$
Category:Primitives involving Logarithm Function
Primitive of Power of x by Power of a x + b

== Theorem ==
=== Decrement of Power of $a x + b$ ===
=== Decrement of Power of $x$ ===
=== Increment of Power of $a x + b$ ===
=== Increment of Power of $x$ ===
Category:Primitive of Power of x by Power of a x + b
Category:Primitives involving a x + b‎
Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b

== Theorem ==
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.83$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.17.$
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b/Proof 1

== Theorem ==
== Proof ==
Let $s \in \Z$. 
Let $u \dfrac {\d v} {\d x} = x^m \paren {a x + b}^n$.
Then:
Let $s$ be selected such that $m + n + 1 - s = 0$.
Then $s = m + n + 1$.
Thus $(2)$ after rearrangement becomes:
:$\dfrac {\d u} {\d x} = \dfrac {x^{m - s} } {m + n + 1} \paren {a x + b}^{n - 1} \paren {\paren {m - s + 1} b}$
Then:
and:
Thus by Integration by Parts:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b/Proof 2

== Theorem ==
== 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 $a := 1, b := 0, p x + q := a x + b$:
Category:Primitive of Power of x by Power of a x + b‎‎
Primitive of Power of x by Power of a x + b/Decrement of Power of x

== Theorem ==
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.83$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.17.$
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Decrement of Power of x/Proof 1

== Theorem ==
== Proof ==
Let $s \in \Z$. 
Let $u \dfrac {\d v} {\d x} = x^m \paren {a x + b}^n$.
Then:
Let $s$ be selected such that $m + n + 1 - s = 0$.
Then $s = m + n + 1$.
Thus $(2)$ after rearrangement becomes:
:$\dfrac {\d u} {\d x} = \dfrac {m b x^{m - 1} \paren {a x + b}^{n - s} } {\paren {m + n + 1} a}$
Then:
and:
Thus by Integration by Parts:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Decrement of Power of x/Proof 2

== Theorem ==
== 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 $p := 1, q := 0, n := m, m := n$:
Category:Primitive of Power of x by Power of a x + b
Primitive of Power of x by Power of a x + b/Increment of Power of a x + b

== Theorem ==
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.83$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.17.$
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Increment of Power of a x + b/Proof 1

== Theorem ==
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
Substituting $n + 1$ for $n$:
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Increment of Power of a x + b/Proof 2

== Theorem ==
== Proof ==
From Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - a \paren {m + n + 2} \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$
Setting $a := 1, b := 0, p x + q := a x + b$:
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Increment of Power of x

== Theorem ==
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
== Proof 1 ==
== Proof 2 ==
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Increment of Power of x/Proof 1

== Theorem ==
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Substituting $m + 1$ for $m$:
Category:Primitive of Power of x by Power of a x + b‎
Primitive of Power of x by Power of a x + b/Increment of Power of x/Proof 2

== Theorem ==
== Proof ==
From Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power:
:$\ds \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac 1 {\paren {n + 1} \paren {b p - a q} } \paren {\paren {a x + b}^{m + 1} \paren {p x + q}^{n + 1} - a \paren {m + n + 2} \int \paren {a x + b}^m \paren {p x + q}^{n + 1} \rd x}$
Setting $p := 1, q := 0, m := n, n := m$:
Category:Primitive of Power of x by Power of a x + b‎‎
Primitive of Power of x by Root of a x + b

== Theorem ==
:$\ds \int x^m \sqrt {a x + b} \rd x = \frac {2 x^m} {\paren {2 m + 3} a} \paren {\sqrt {a x + b} }^3 - \frac {2 m b} {\paren {2 m + 3} a} \int x^{m - 1} \sqrt{a x + b} \rd x$
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Putting $n := \dfrac 1 2$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.96$
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.13.$
Category:Primitives involving Root of a x + b
Primitive of Power of x by Sine of a x

== Theorem ==
:$\ds \int x^m \sin a x \rd x = \frac {-x^m \cos a x} a + \frac {m x^{m - 1} \sin a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \int x^{m - 2} \sin a x \rd x$
== Proof ==
=== Lemma ===
From :
:$(1): \quad \ds \int x^{m - 1} \cos a x \rd x = \frac {x^{m - 1} \sin a x} a - \frac {m - 1} a \int x^{m - 2} \sin a x \rd x$
So:
== Also see ==
* Primitive of $x^m \cos a x$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.364$
Category:Primitive of Power of x by Sine of a x
Category:Primitives involving Sine Function
Primitive of Power of x by Sine of a x/Lemma

== Theorem ==
:$\ds \int x^m \sin a x \rd x = \frac {- x^m \cos a x} a + \frac m a \int x^{m - 1} \cos a x \rd x$
== 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 $x^m \cos a x$
== Sources ==
* : Back endpapers: A Brief Table of Integrals: $80$.
Category:Primitive of Power of x by Sine of a x
Primitive of Power of x less one over Power of x minus Power of a

== Theorem ==
:$\ds \int \frac {x^{n - 1} \rd x} {x^n - a^n} = \frac 1 n \ln \size {x^n - a^n} + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.331$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Power of x less one over Power of x plus Power of a

== Theorem ==
:$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.326$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Power of x over Even Power of x minus Even Power of a

== Theorem ==
where $0 < p \le 2 m$.
== Proof ==
The integrand is a rational function.
It has simple poles at $x = \omega_k a$ where $\omega_k = e^{\pi i k /m} $, $k = 0, 1, \ldots, 2 m - 1$ are the $2m$'th roots of unity:
Its residue at $x = \omega_k$ is then:
:$\Res f {\omega_k a} = \dfrac {\paren {\omega_k a}^{p - 2m} } {2 m}$
so that:
:$\ds \map f x = \sum_{k \mathop = 0}^{2 m - 1} \dfrac {\paren {\omega_k a}^{p - 2m} } {2 m \paren {x - \omega_k a} }$
which has a primitive:
:$\ds \map F x = \sum_{k \mathop = 0}^{2 m - 1} \dfrac {\paren {\omega_k a}^{p - 2m} } {2 m} \map \ln {x - \omega_k a}$
For positive real $x, a \in \R_{>0}$ we can write:
:$\map \ln {x - \omega_k a} = \dfrac 1 2 \map \ln {x^2 - 2 a x \map \cos {\dfrac {2 \pi k} m} + a^2} - i \map \arctan {\dfrac {a \map \sin {\dfrac {2 \pi k} m} } {x - a \map \cos {\dfrac {2 \pi k} m} } }$
:$\paren {\omega_k a}^{p - 2 m} = a^{p - 2 m} \paren {\map \cos {\dfrac {p k \pi} m} + i \map \sin {\dfrac {p k \pi} m} }$
and the formula follows.
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.336$
* : $\S 17$: Tables of Special Indefinite Integrals: $(16)$ Integrals Involving $x^n \pm a^n$: $17.16.12.$
* 
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Power of x over Even Power of x plus Even Power of a

== Theorem ==
where $0 < p \le 2 m$.
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.335$
* : $\S 17$: Tables of Special Indefinite Integrals: $(16)$ Integrals Involving $x^n \pm a^n$: $17.16.11.$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Power of x over Odd Power of x minus Odd Power of a

== Theorem ==
where $0 < p \le 2 m + 1$.
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.338$
* : $\S 17$: Tables of Special Indefinite Integrals: $(16)$ Integrals Involving $x^n \pm a^n$: $17.16.14.$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Power of x over Odd Power of x plus Odd Power of a

== Theorem ==
where $0 < p \le 2 m + 1$.
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.337$
* : $\S 17$: Tables of Special Indefinite Integrals: $(16)$ Integrals Involving $x^n \pm a^n$: $17.16.13.$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Power of x over Power of Power of x minus Power of a

== Theorem ==
:$\ds \int \frac {x^m \rd x} {\paren {x^n - a^n}^r} = a^n \int \frac {x^{m - n} \rd x} {\paren {x^n - a^n}^r} + \int \frac {x^{m - n} \rd x} {\paren {x^n - a^n}^{r - 1} }$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.332$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Power of x over Power of Power of x plus Power of a

== Theorem ==
:$\ds \int \frac {x^m \rd x} {\paren {x^n + a^n}^r} = \int \frac {x^{m - n} \rd x} {\paren {x^n + a^n}^{r - 1} } - a^n \int \frac {x^{m - n} \rd x} {\paren {x^n + a^n}^r}$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.327$
Category:Primitives involving Power of x plus or minus Power of a