nyuuzyou/wikis huggingface dataset:
Primitive of x over Root of a squared minus x squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\sqrt {a^2 - x^2} } = -\sqrt {a^2 - x^2} + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$
* Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.238$
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
* : $\S 17$: Tables of Special Indefinite Integrals: $(11)$ Integrals Involving $\sqrt {a^2 - x^2}$: $17.11.2.$
Category:Primitives involving Root of a squared minus x squared
Primitive of x over Root of a squared minus x squared cubed

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {\sqrt {a^2 - x^2} }^3} = \frac 1 {\sqrt {a^2 - x^2} } + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac x {\paren {\sqrt {x^2 - a^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.252$
Category:Primitives involving Root of a squared minus x squared
Primitive of x over Root of a x + b

== Theorem ==
:$\ds \int \frac {x \rd x} {\sqrt {a x + b} } = \frac {2 \paren {a x - 2 b} \sqrt {a x + b} } {3 a^2}$
== Proof ==
Let:
Thus:
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.85$
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.2.$
Category:Primitives involving Root of a x + b
Primitive of x over Root of a x + b by Root of p x + q

== Theorem ==
:$\ds \int \frac {x \rd x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } {a p} - \frac {b p + a q} {2 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
== Proof ==
Let:
Let $c^2 = \dfrac {b p - a q} p$.
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $14.121$
* : $\S 17$: Tables of Special Indefinite Integrals: $(5)$ Integrals Involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $17.5.2.$
Category:Primitives involving Root of a x + b and Root of p x + q
Primitive of x over Root of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} } = \frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
== Proof ==
First:
Then:
== Examples ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.39$
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.281$
* : $\S 17$: Tables of Special Indefinite Integrals: $(13)$ Integrals Involving $\sqrt {a x^2 + b x + c}$: $17.13.2.$
Category:Primitive of x over Root of a x squared plus b x plus c
Category:Primitives involving Root of a x squared plus b x plus c
Primitive of x over Root of a x squared plus b x plus c/Examples

== Examples of Use of Primitive of $\dfrac x {\sqrt {a x^2 + b x + c} }$ ==
=== Primitive of $\dfrac x {\sqrt {x^2 + 4 x + 5} }$ ===
Category:Examples of Use of Primitive of x over Root of a x squared plus b x plus c
Primitive of x over Root of a x squared plus b x plus c/Examples/x over Root of x^2 + 4 x + 5

== Examples of Use of Primitive of $\dfrac x {\sqrt {a x^2 + b x + c} }$ ==
:$\ds \int \dfrac {x \rd x} {\sqrt {x^2 + 4 x + 5} } = \sqrt {x^2 + 4 x + 5} - 2 \map \ln {x + 2 + \sqrt {x^2 + 4 x + 5} } + C$
== Proof ==
From Primitive of $\dfrac x {\sqrt {a x^2 + b x + c} }$ with:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $18$.
Category:Examples of Use of Primitive of x over Root of a x squared plus b x plus c
Primitive of x over Root of x squared minus a squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\sqrt {x^2 - a^2} } = \sqrt {x^2 - a^2} + C$
for $\size x > a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$
* Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.210$
* : $\S 17$: Tables of Special Indefinite Integrals: $(10)$ Integrals Involving $\sqrt {x^2 - a^2}$: $17.10.1.$
Category:Primitives involving Root of x squared minus a squared
Primitive of x over Root of x squared minus a squared cubed

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3} = \frac {-1} {\sqrt {x^2 - a^2} } + C$
for $\size x > a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac x {\paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.224$
Category:Primitives involving Root of x squared minus a squared
Primitive of x over Root of x squared plus a squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\sqrt {x^2 + a^2} } = \sqrt {x^2 + a^2} + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$
* Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.183$
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
* : $\S 17$: Tables of Special Indefinite Integrals: $(9)$ Integrals Involving $\sqrt {x^2 + a^2}$: $17.9.2.$
Category:Primitives involving Root of x squared plus a squared
Primitive of x over Root of x squared plus a squared cubed

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {\sqrt {x^2 + a^2} }^3} = \frac {-1} {\sqrt {x^2 + a^2} } + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\paren {\sqrt {x^2 - a^2} }^3}$
* Primitive of $\dfrac x {\paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.197$
Category:Primitives involving Root of x squared plus a squared
Primitive of x over Sine of a x

== Theorem ==
where $B_n$ denotes the $n$th Bernoulli number.
== Proof ==
== Also see ==
* Primitive of $\dfrac x {\cos a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.346$
Category:Primitives involving Sine Function
Primitive of x over a squared minus x squared

== Theorem ==
:$\ds \int \frac {x \rd x} {a^2 - x^2} = -\frac 1 2 \map \ln {a^2 - x^2} + C$
for $x^2 < a^2$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {x^2 - a^2}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.164$
* : $\S 17$: Tables of Special Indefinite Integrals: $(8)$ Integrals Involving $x^2 - a^2$, $x^2 < a^2$: $17.8.2.$
Category:Primitives involving a squared minus x squared
Primitive of x over a squared minus x squared squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {a^2 - x^2}^2} = \frac 1 {2 \paren {a^2 - x^2} } + C$
for $x^2 < a^2$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\paren {x^2 - a^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.171$
Category:Primitives involving a squared minus x squared
Primitive of x over a x + b

== Theorem ==
:$\ds \int \frac {x \rd x} {a x + b} = \frac x a - \frac b {a^2} \ln \size {a x + b} + C$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Primitive of $x$ by $\paren {a x + b}^n$ for general $n$
* Primitive of $x$ over $\paren {a x + b}^2$ for the case when $n = -2$
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.60$
* : Front endpapers: A Brief Table of Integrals: $8$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.2.$
Category:Primitive of x over a x + b‎
Category:Primitives involving a x + b
Primitive of x over a x + b/Proof 2

== Theorem ==
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Let $m = 1$ and $n = -1$.
Then:
Category:Primitive of x over a x + b‎
Primitive of x over a x + b by p x + q

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {a x + b} \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac b a \ln \size {a x + b} - \frac q p \ln \size {p x + q} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.106$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.2.$
Category:Primitive of x over a x + b by p x + q
Category:Primitives involving a x + b and p x + q
Primitive of x over a x + b by p x + q/Partial Fraction Expansion

== Lemma for Primitive of $\frac x {\paren {a x + b} \paren {p x + q} }$ ==
:$\dfrac x {\paren {a x + b} \paren {p x + q} } \equiv \dfrac b {\paren {b p - a q} \paren {a x + b} } - \dfrac q {\paren {b p - a q} \paren {p x + q} }$
== Proof ==
Setting $p x + q = 0$ in $(1)$:
Setting $a x + b = 0$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of x over a x + b by p x + q
Category:Primitives involving a x + b and p x + q/Lemmata
Primitive of x over a x + b cubed

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {a x + b}^3} = \frac {-1} {a^2 \paren {a x + b} } + \frac b {2 a^2 \paren {a x + b}^2} + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.74$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.12.$
Category:Primitive of x over a x + b cubed
Category:Primitives involving a x + b‎
Primitive of x over a x + b cubed/Proof 2

== Theorem ==
== Proof ==
From Primitive of $x$ by Power of $a x + b$:
:$\ds \int x \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 2} } {\paren {n + 2} a^2} - \frac {b \paren {a x + b}^{n + 1} } {\paren {n + 1} a^2} + C$
where $n \ne - 1$ and $n \ne - 2$.
The result follows by setting $n = -3$.
Category:Primitive of x over a x + b cubed
Primitive of x over a x + b squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {a x + b}^2} = \frac b {a^2 \paren {a x + b} } + \frac 1 {a^2} \ln \size {a x + b} + C$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Also see ==
* Primitive of $x$ by $\paren {a x + b}^n$ for general $n$
* Primitive of $x$ over $a x + b$ for the case when $n = -1$
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.67$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.7.$
Category:Primitive of x over a x + b squared
Category:Primitives involving a x + b
Primitive of x over a x + b squared/Also presented as

== Primitive of $x$ over $\paren {a x + b}^2$: Also presented as ==
This result is also seen presented in the form:
:$\ds \int \frac {x \rd x} {\paren {a x + b}^2} = \frac 1 {a^2} \paren {\ln \size {a x + b} + \frac b {a x + b} } + C$
== Also see ==
* Primitive of $x$ by $\paren {a x + b}^n$ for general $n$
* Primitive of $x$ over $a x + b$ for the case when $n = -1$
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $9$.
Category:Primitive of x over a x + b squared
Primitive of x over a x + b squared by p x + q

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C$
=== Corollary ===
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.108$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.4.$
Category:Primitive of x over a x + b squared by p x + q
Category:Primitives involving a x + b and p x + q
Primitive of x over a x + b squared by p x + q/Corollary

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } + \frac x {a x + b} } + C$
== Proof ==
Category:Primitive of x over a x + b squared by p x + q
Category:Primitives involving a x + b and p x + q
Primitive of x over a x + b squared by p x + q/Partial Fraction Expansion

== Lemma for Primitive of $\frac x {\paren {a x + b}^2 \paren {p x + q} }$ ==
:$\dfrac x {\paren {a x + b}^2 \paren {p x + q} } \equiv \dfrac 1 {b p - a q} \paren {\dfrac {a q} {\paren {b p - a q} \paren {a x + b} } + \dfrac b {\paren {a x + b}^2} - \dfrac {p q} {\paren {b p - a q} \paren {p x + q} } }$
== Proof ==
Setting $a x + b = 0$ in $(1)$:
Setting $p x + q = 0$ in $(1)$:
Equating $2$nd powers of $x$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of x over a x + b squared by p x + q
Category:Primitives involving a x + b and p x + q/Lemmata
Primitive of x over a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x \rd x} {a x^2 + b x + c} = \frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c}$
== Proof ==
First note that by Derivative of Power:
:$(1): \quad \map {\dfrac \d {\d x} } {a x^2 + b x + c} = 2 a x + b$
Then:
== Examples ==
== Also see ==
* Primitive of $\dfrac 1 {a x^2 + b x + c}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.19$
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.266$
* : $\S 17$: Tables of Special Indefinite Integrals: $(12)$ Integrals Involving $a x^2 + b x + c$: $17.12.2.$
Category:Primitive of x over a x squared plus b x plus c
Category:Primitives involving a x squared plus b x plus c
Primitive of x over a x squared plus b x plus c/Examples/x over x^2 + 4 x + 5

== Examples of Use of Primitive of $\dfrac x {a x^2 + b x + c}$ ==
:$\ds \int \dfrac {x \rd x} {x^2 + 4 x + 5} = \frac 1 2 \map \ln {x^2 + 4 x + 5} - 2 \map \arctan {x + 2} + C$
== Proof ==
From Primitive of $\dfrac x {a x^2 + b x + c}$ with:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $17$.
Category:Examples of Use of Primitive of x over a x squared plus b x plus c
Primitive of x over square of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2} = \frac {-\paren {b x + 2 c} } {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$
== Proof ==
Let:
Then:
== Sources ==
* : $\S 14$: Integrals involving $a x^2 + b x + c$: $14.273$
Category:Primitives involving a x squared plus b x plus c
Primitive of x over x cubed plus a cubed

== Theorem ==
:$\ds \int \frac {x \rd x} {x^3 + a^3} = \frac 1 {6 a} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.300$
* : $\S 17$: Tables of Special Indefinite Integrals: $(14)$ Integrals Involving $x^3 + a^3$: $17.14.2.$
Category:Primitives involving x cubed plus a cubed
Primitive of x over x cubed plus a cubed squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {x^3 + a^3}^2} = \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {18 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {3 a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$ 
== Proof ==
First a lemma:
=== Lemma ===
Then:
== Sources ==
* : $\S 14$: Integrals involving $x^3 + a^3$: $14.305$
Category:Primitives involving x cubed plus a cubed
Category:Primitive of x over x cubed plus a cubed squared
Primitive of x over x cubed plus a cubed squared/Lemma

== Primitive of x over x cubed plus a cubed squared: Lemma ==
:$\ds \int \frac {x \rd x} {\paren {x^3 + a^3}^2} = \frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} }$ 
== Proof ==
Then from Primitive of $\dfrac {\d x} {\paren {a x + b}^m \paren {p x + q}^n}$:
:$\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$
Here we have $a = 1, b = 0, m = \dfrac 1 3, p = 1, q = a^3, n = 2$.
So:
Category:Primitive of x over x cubed plus a cubed squared
Category:Primitives involving x cubed plus a cubed/Lemmata
Primitive of x over x fourth minus a fourth

== Theorem ==
:$\ds \int \frac {x \rd x} {x^4 - a^4} = \frac 1 {4 a^2} \ln \size {\frac {x^2 - a^2} {x^2 + a^2} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.319$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of x over x fourth plus a fourth

== Theorem ==
:$\ds \int \frac {x \rd x} {x^4 + a^4} = \frac 1 {2 a^2} \arctan \frac {x^2} {a^2}$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.312$
* : $\S 17$: Tables of Special Indefinite Integrals: $(15)$ Integrals Involving $x^4 \pm a^4$: $17.15.2.$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of x over x squared minus a squared

== Theorem ==
:$\ds \int \frac {x \rd x} {x^2 - a^2} = \frac 1 2 \map \ln {x^2 - a^2} + C$
for $x^2 > a^2$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {a^2 - x^2}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.145$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.2.$
Category:Primitives involving x squared minus a squared
Primitive of x over x squared minus a squared squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {x^2 - a^2}^2} = \frac {-1} {2 \paren {x^2 - a^2} } + C$
for $x^2 > a^2$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac x {\paren {a^2 - x^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.152$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.9.$
Category:Primitives involving x squared minus a squared
Primitive of x over x squared plus a squared

== Theorem ==
:$\ds \int \frac {x \rd x} {x^2 + a^2} = \frac 1 2 \map \ln {x^2 + a^2} + C$
where $a$ is a non-zero constant.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.126$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.2.$
Category:Primitive of x over x squared plus a squared
Category:Primitives involving x squared plus a squared
Primitive of x over x squared plus a squared/Corollary

== Theorem ==
:$\ds \int \frac {x \rd x} {a^2 + b^2 x^2} = \frac 1 {2 b^2} \map \ln {a^2 + b^2 x^2} + C$
where $a$ is a non-zero constant.
== Proof ==
Let $z = b x$.
Then:
:$\dfrac {\d x} {\d z} = \dfrac 1 b$
Hence:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.22$
Category:Primitive of x over x squared plus a squared
Primitive of x over x squared plus a squared/Proof 2

== Theorem ==
== Proof ==
From Primitive of Power of x less one over Power of x plus Power of a:
:$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$
So:
Category:Primitive of x over x squared plus a squared
Primitive of x over x squared plus a squared squared

== Theorem ==
:$\ds \int \frac {x \rd x} {\paren {x^2 + a^2}^2} = -\frac 1 {2 \paren {x^2 + a^2} } + C$
== Proof ==
Let:
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.133$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.9.$
Category:Primitives involving x squared plus a squared
Primitive of x sixth by Cosine of a x

== Theorem ==
:$\ds \int x^6 \cos a x \rd x = \frac {\sin a x} a x^6 + \frac {6 \cos a x} {a^2} x^5 - \frac {30 \sin a x} {a^3} x^4 - \frac {120 \cos a x} {a^4} x^3 + \frac {360 \sin a x} {a^5} x^2 + \frac {720 \cos a x} {a^6} x - \frac {720 \sin a x} {a^7} + C$
where $C$ is an arbitrary constant.
== Proof ==
== Also see ==
* Primitive of $x^4 \cos a x$
* Primitive of $x^2 \cos a x$
* Primitive of $x^2 \sin a x$
Category:Primitives involving Cosine Function
Primitive of x squared by Arccosecant of x over a

== Theorem ==
:$\ds \int x^2 \arccsc \frac x a \rd x = \begin {cases}
\dfrac {x^3} 3 \arccsc \dfrac x a - \dfrac {a x \sqrt{x^2 - a^2} } 6 - \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac {x^3} 3 \arccsc \dfrac x a + \dfrac {a x \sqrt{x^2 - a^2} } 6 + \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end {cases}$
== 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 $\arccsc \dfrac x a$ 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 $x^2 \arcsin \dfrac x a$
* Primitive of $x^2 \arccos \dfrac x a$
* Primitive of $x^2 \arctan \dfrac x a$
* Primitive of $x^2 \arccot \dfrac x a$
* Primitive of $x^2 \arcsec \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.500$
Category:Primitives involving Inverse Cosecant Function
Primitive of x squared by Arccosine of x over a

== Theorem ==
:$\ds \int x^2 \arccos \frac x a \rd x = \frac {x^3} 3 \arccos \frac x a - \frac {\paren {x^2 + 2 a^2} \sqrt {a^2 - x^2} } 9 + C$
== 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 $x^2 \arcsin \dfrac x a$
* Primitive of $x^2 \arctan \dfrac x a$
* Primitive of $x^2 \arccot \dfrac x a$
* Primitive of $x^2 \arcsec \dfrac x a$
* Primitive of $x^2 \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.479$
Category:Primitives involving Inverse Cosine Function
Primitive of x squared by Arccotangent of x over a

== Theorem ==
:$\ds \int x^2 \arccot \frac x a \rd x = \frac {x^3} 3 \arccot \frac x a + \frac {a x^2} 6 - \frac {a^3} 6 \map \ln {x^2 + a^2} + C$
== 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 $x^2 \arcsin \dfrac x a$
* Primitive of $x^2 \arccos \dfrac x a$
* Primitive of $x^2 \arctan \dfrac x a$
* Primitive of $x^2 \arcsec \dfrac x a$
* Primitive of $x^2 \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.490$
Category:Primitives involving Inverse Cotangent Function
Primitive of x squared by Arcsecant of x over a

== Theorem ==
:$\ds \int x^2 \arcsec \frac x a \rd x = \begin{cases}
\dfrac {x^3} 3 \arcsec \dfrac x a - \dfrac {a x \sqrt {x^2 - a^2} } 6 - \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {x^3} 3 \arcsec \dfrac x a + \dfrac {a x \sqrt {x^2 - a^2} } 6 + \dfrac {a^3} 6 \map \ln {x + \sqrt {x^2 - a^2} } + C & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end{cases}$
== 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 $\arcsec \dfrac x a$ 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 $x^2 \arcsin \dfrac x a$
* Primitive of $x^2 \arccos \dfrac x a$
* Primitive of $x^2 \arctan \dfrac x a$
* Primitive of $x^2 \arccot \dfrac x a$
* Primitive of $x^2 \arccsc \dfrac x a$
== Sources ==
* : $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.495$
Category:Primitives involving Inverse Secant Function