nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of x squared by x fourth minus a fourth

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^4 - a^4} } = \frac 1 {a^4 x} + \frac 1 {4 a^5} \ln \size {\frac {x - a} {x + a} } + \frac 1 {2 a^5} \arctan \frac x a + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.323$
* : $\S 17$: Tables of Special Indefinite Integrals: $(15)$ Integrals Involving $x^4 \pm a^4$: $17.15.13.$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x squared by x fourth minus a fourth/Mistake

== Source Work ==
:Chapter $17$: Tables of Special Indefinite Integrals
::$(15)$ Integrals Involving $x^4 \pm a^4$: $17.15.13$
== Mistake ==
:$\ds \int \frac {\d x} {x^3 \paren {x^4 - a^4} } = \frac 1 {a^4 x} + \frac 1 {4 a^5} \ln \size {\frac {x - a} {x + a} } + \frac 1 {2 a^5} \arctan \frac x a + C$
== Correction ==
That should read:
:$\ds \int \frac {\d x} {x^{\color {red} 2} \paren {x^4 - a^4} } = \frac 1 {a^4 x} + \frac 1 {4 a^5} \ln \size {\frac {x - a} {x + a} } + \frac 1 {2 a^5} \arctan \frac x a + C$
where the correction is implemented in .
== Sources ==
* : $\S 17$: Tables of Special Indefinite Integrals: $(15)$ Integrals Involving $x^4 \pm a^4$: $17.15.13.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Primitive of Reciprocal of x squared by x fourth plus a fourth

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^4 + a^4} } = \frac {-1} {a^4 x} - \frac {-1} {4 a^5 \sqrt 2} \map \ln {\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} } + \frac 1 {2 a^5 \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} }$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^4 \pm a^4$: $14.316$
Category:Primitives involving x to the fourth plus or minus a to the fourth
Primitive of Reciprocal of x squared by x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^2 - a^2} } = \frac 1 {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {x - a} {x + a} } + C$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \paren {a^2 - x^2} }$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.149$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.6.$
Category:Primitive of Reciprocal of x squared by x squared minus a squared
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of x squared by x squared minus a squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^2 \left({x^2 - a^2}\right)$ ==
:$\dfrac 1 {x^2 \paren {x^2 - a^2} } \equiv \dfrac 1 {a^2 \paren {x^2 - a^2} } - \dfrac 1 {a^2 x^2}$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x squared by x squared minus a squared
Category:Primitives involving x squared minus a squared/Lemmata
Primitive of Reciprocal of x squared by x squared minus a squared squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^2 - a^2}^2} = \frac {-1} {a^4 x} - \frac x {2 a^4 \paren {x^2 - a^2} } + \frac 3 {4 a^5} \map \ln {\frac {x + a} {x - a} } + C$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 \paren {a^2 - x^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.156$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.13.$
Category:Primitive of Reciprocal of x squared by x squared minus a squared squared
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of x squared by x squared minus a squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^2 \paren {x^2 - a^2}^2$ ==
:$\dfrac 1 {x^2 \paren {x^2 - a^2}^2} \equiv \dfrac 1 {a^4 x^2} + \dfrac 3 {4 a^5 \paren {x + a} } - \dfrac 3 {4 a^5 \paren {x - a} } + \dfrac 1 {4 a^4 \paren {x + a}^2} + \dfrac 1 {4 a^4 \paren {x - a}^2}$
== Proof ==
Setting $x = a$ in $(1)$:
Setting $x = -a$ in $(1)$:
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^5$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x squared by x squared minus a squared squared
Category:Primitives involving x squared minus a squared/Lemmata
Primitive of Reciprocal of x squared by x squared plus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^2 + a^2} } = -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.130$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.6.$
Category:Primitive of Reciprocal of x squared by x squared plus a squared
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of x squared by x squared plus a squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^2 \paren {x^2 + a^2}$ ==
:$\dfrac 1 {x^2 \paren {x^2 + a^2} } \equiv \dfrac 1 {a^2 x^2} - \dfrac 1 {a^2 \paren {x^2 + a^2} }$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x squared by x squared plus a squared
Category:Primitives involving x squared plus a squared/Lemmata
Primitive of Reciprocal of x squared by x squared plus a squared squared

== Theorem ==
:$\ds \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2} = -\frac 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C$
== Proof ==
Let:
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.137$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.13.$
Category:Primitive of Reciprocal of x squared by x squared plus a squared squared
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of x squared by x squared plus a squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $x^2 \paren {x^2 + a^2}^2$ ==
:$\dfrac 1 {x^2 \paren {x^2 + a^2}^2} \equiv -\dfrac 1 {a^4 x^2} - \dfrac 1 {a^4 \paren {x^2 + a^2} } - \dfrac 1 {a^2 \paren {x^2 + a^2}^2}$
== Proof ==
Setting $x = 0$ in $(1)$:
Equating coefficients of $x^4$ in $(1)$:
Equating coefficients of $x$ in $(1)$:
Equating coefficients of $x^5$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x squared by x squared plus a squared squared
Category:Primitives involving x squared plus a squared/Lemmata
Primitive of Reciprocal of x squared minus a squared

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
=== Inverse Hyperbolic Function Form ===
=== $1$st Logarithm Form ===
=== $2$nd Logarithm Form ===
== Also see ==
* Primitive of $\dfrac 1 {x^2 + a^2}$
* Primitive of $\dfrac 1 {a^2 - x^2}$
== Sources ==
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(viii)}$
* : $\S 14$: General Rules of Integration: $14.40$
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.144$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.40.$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.1.$
Category:Primitive of Reciprocal of x squared minus a squared
Category:Primitives involving x squared minus a squared
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Cotangent Form

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $\size x > a$.
Then:
:$\ds \int \frac {\d x} {x^2 - a^2} = -\frac 1 a \coth^{-1} {\frac x a} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 - a^2}$: $\tanh^{-1}$ form for the case $a^2 > x^2$
== Sources ==
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(viii)}$
* : $\S 14$: General Rules of Integration: $14.40$
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.144$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.40.$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.1.$
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Function Form

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
$\quad \ds \int \dfrac {\d x} {x^2 - a^2} = \begin {cases} -\dfrac 1 a \tanh^{-1} \dfrac x a + C & : \size x < a \\
& \\
-\dfrac 1 a \coth^{-1} \dfrac x a + C & : \size x > a \\
& \\
\text {undefined} & : x = a \end {cases}$
== Proof ==
First note that if $x = a$ then $a^2 - x^2 = 0$ and so $\dfrac 1 {x^2 - a^2}$ is undefined.
=== Proof for $\size x < a$ ===
=== Proof for $\size x > a$ ===
== Sources ==
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Tangent Form

== Theorem ==
:$\ds \int \frac {\d x} {x^2 - a^2} = -\frac 1 a \tanh^{-1} {\frac x a} + C$
where $\size x < a$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 - a^2}$: $\coth^{-1}$ form for the case $\size z > a$
== Sources ==
* : Appendix $\text I$: Table of Indefinite Integrals $13$.
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$ such that $\size x \ne a$.
=== $1$st Logarithm Form ===
=== $2$nd Logarithm Form ===
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Lemma

== Lemma ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Then:
:$\map \ln {\dfrac {x - a} {x + a} }$ is defined $\size x > a$
:$\map \ln {\dfrac {a - x} {a + x} }$ is defined $\size x < a$
== Proof ==
We have that the real natural logarithm is defined only on the strictly positive real numbers.
Hence:
:$\map \ln {\dfrac {x - a} {x + a} }$ is defined $\dfrac {x - a} {x + a} > 0$
:$\map \ln {\dfrac {a - x} {a + x} }$ is defined $\dfrac {a - x} {a + x} > 0$
First we note that if $\size x = a$, then either the numerator or denominator of the arguments of the logarithm functions in question are either $0$ or undefined.
Hence the expressions have no meaning unless $\size x \ne a$.
The following table indicates whether each of $x + a$, $a - x$ and $x - a$ are positive $(+)$ or negative $(-)$ on the domains in question.
$\begin {array} {c|ccc|cc}
& x + a & a - x & x - a & \dfrac {x - a} {x + a} & \dfrac {a - x} {a + x} \\
\hline 
a < x & + & - & + & + & - \\
0 < x < a & + & + & - & - & + \\
-a < x < 0 & + & + & - & - & + \\
x < -a & - & + & - & + & - \\
\end {array}$
Hence:
:$\map \ln {\dfrac {x - a} {x + a} }$ is defined $x > a$ or $x < -a$
:$\map \ln {\dfrac {a - x} {a + x} }$ is defined $-a < x < a$
as we were required to show.
Category:Logarithms
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$.
Then:
$\quad \ds \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a\\
& \\
\dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\
& \\
\text {undefined} & : \size x = a \end {cases}$
== Proof ==
=== Case where $\size x < a$ ===
=== Case where $\size x > a$ ===
== Also see ==
* Primitive of $\dfrac 1 {x^2 + a^2}$
* Primitive of $\dfrac 1 {a^2 - x^2}$: Logarithm Form
== Sources ==
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x greater than a

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $\size x > a$.
Then:
:$\ds \int \frac {\d x} {x^2 - a^2} = \dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C$
== Proof ==
Let $\size x > a$.
Then:
== Sources ==
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(viii)}$
* : $\S 14$: General Rules of Integration: $14.40$
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.144$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.40.$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.1.$
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x less than a

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $\size x < a$.
Then:
:$\ds \int \frac {\d x} {x^2 - a^2} = \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C$
== Proof ==
Let $\size x < a$.
Then:
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$ such that $\size x \ne a$.
:$\ds \int \frac {\d x} {x^2 - a^2} = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text {III}$: Integration: Integration
* : Appendix $\text I$: Table of Indefinite Integrals $13$.
* : Chapter $25$: Fundamental Integration Formulas: $21$.
* : Chapter $5$. Integrals: Integrals of Special Functions: $28$
* : Appendix $7$: Integrals
* : Appendix $8$: Integrals
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 1

== Theorem ==
== Proof ==
Let $x > a$.
Then:
Let $x < -a$.
Let $z = -x$.
Then:
:$\d x = -\d z$
and we then have:
The result follows.
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 2/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $\paren {x^2 - a^2}$ ==
:$\dfrac 1 {x^2 - a^2} \equiv \dfrac 1 {2 a \paren {x - a} } - \dfrac 1 {2 a \paren {x + a} }$
== Proof ==
Setting $x = a$ in $(1)$:
Setting $x = -a$ in $(1)$:
Summarising:
Hence the result.
Category:Primitives involving x squared minus a squared/Lemmata
Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 3

== Theorem ==
== Proof ==
From the $1$st logarithm form:
$\quad \ds \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a \\
& \\
\dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\
& \\
\text {undefined} & : \size x = a \end {cases}$
From Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma:
:$\map \ln {\dfrac {x - a} {x + a} }$ is defined $\size x > a$
:$\map \ln {\dfrac {a - x} {a + x} }$ is defined $\size x < a$
Let $\size x > a$.
Then $\map \ln {\dfrac {x - a} {x + a} }$ is defined.
We have that:
So the result holds for $\size x > a$.
Let $\size x < a$.
Then $\map \ln {\dfrac {a - x} {a + x} }$ is defined.
We have:
We have that:
The result follows.
Category:Primitive of Reciprocal of x squared minus a squared
Primitive of Reciprocal of x squared minus a squared squared

== Theorem ==
:$\ds \int \frac {\d x} {\paren {x^2 - a^2}^2} = \frac {-x} {2 a^2 \paren {x^2 - a^2} } + \frac 1 {4 a^3} \map \ln {\frac {x + a} {x - a} } + C$
for $x^2 > a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {\paren {a^2 - x^2}^2}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.25$
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.151$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.8.$
Category:Primitive of Reciprocal of x squared minus a squared squared
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of x squared minus a squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $\paren {x^2 - a^2}^2$ ==
:$\dfrac 1 {\paren {x^2 - a^2}^2} \equiv \dfrac 1 {4 a^3 \paren {x + a} } - \dfrac 1 {4 a^3 \paren {x - a} } + \dfrac 1 {4 a^2 \paren {x + a}^2} + \dfrac 1 {4 a^2 \paren {x - a}^2}$
== Proof ==
Setting $x = a$ in $(1)$:
Setting $x = -a$ in $(1)$:
Equating coefficients of $x^3$ in $(1)$:
Equating coefficients of $x^2$ in $(1)$:
Summarising:
Hence the result.
Category:Primitive of Reciprocal of x squared minus a squared squared
Category:Primitives involving x squared minus a squared/Lemmata
Primitive of Reciprocal of x squared plus a squared/Arccotangent Form

== Theorem ==
:$\ds \int \frac {\d x} {x^2 + a^2} = -\frac 1 a \arccot \frac x a + C$
where $a$ is a non-zero constant.
== Proof ==
== Sources ==
* : Appendix $\text I$: Table of Indefinite Integrals $12$.
* : Chapter $5$. Integrals: Integrals of Special Functions: $27$
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
Category:Primitive of Reciprocal of x squared plus a squared
Primitive of Reciprocal of x squared plus a squared/Arctangent Form

== Theorem ==
:$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$
where $a$ is a non-zero constant.
=== Corollary $1$ ===
=== Corollary $2$ ===
== Proof $1$ ==
== Proof $2$ ==
== Proof $3$ ==
== Proof $4$ ==
== Also see ==
* Derivative of Arctangent Function
* Primitive of $\dfrac 1 {x^2 - a^2}$
* Primitive of $\dfrac 1 {a^2 - x^2}$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $13$.
* : Chapter $\text {III}$: Integration: Integration
* : $\text {II}$. Calculus: Integration: Algebraic Integration: $\text {II}$.
* : $\text {II}$. Calculus: Integration
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(vi)}$
* : $\S 14$: General Rules of Integration: $14.39$
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.125$
* : Front endpapers: A Brief Table of Integrals: $16$.
* : Appendix $\text I$: Table of Indefinite Integrals $12$.
* : Chapter $25$: Fundamental Integration Formulas: $19$.
* : Chapter $5$. Integrals: Integrals of Special Functions: $27$
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
* : $9$ Sums and Integrals: $9.8$ Standard Integrals
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.39.$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.1.$
* : Appendix $7$: Integrals
* : Appendix $8$: Integrals
Category:Primitive of Reciprocal of x squared plus a squared
Category:Expressions whose Primitives are Inverse Trigonometric Functions
Category:Primitives involving x squared plus a squared
Category:Arctangent Function
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Corollary 2

== Corollary to Primitive of Reciprocal of x squared plus a squared ==
:$\ds \int \frac {\d x} {a^2 + b^2 x^2} = \frac 1 {a b} \arctan \frac {b x} a + C$
where $a$ and $b$ are non-zero constants.
== 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: $3.3.21$
Category:Primitive of Reciprocal of x squared plus a squared
Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 1

== Theorem ==
== Proof ==
Let:
:$a \tan \theta = x$
for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Shape of Tangent Function, this substitution is valid for all real $x$.
Then:
As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
:$\tan \theta = \dfrac x a \iff \theta = \arctan \dfrac x a$
Thus:
:$\ds \int \frac 1 {x^2 + a^2} \rd x = \frac 1 a \arctan \frac x a + C$
When $a = 0$, both $\dfrac x a$ and $\dfrac 1 a$ are undefined.
However, consider the limit of the above primitive as $a \to 0$:
This corresponds with the result:
:$\ds \int \frac 1 {x^2} \rd x = \frac {-1} x + C$
which follows from Primitive of Power.
== Sources ==
* 
Category:Primitive of Reciprocal of x squared plus a squared
Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 2

== Theorem ==
== Proof ==
We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$.
Thus from Primitive of $\dfrac 1 {a x^2 + b x + c}$ for $b^2 - 4 a c > 0$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
setting $a := 1, b := 0, c := a^2$:
Category:Primitive of Reciprocal of x squared plus a squared
Primitive of Reciprocal of x squared plus a squared/Examples/16 + x^2

== Example of Use of Primitive of $\dfrac 1 {x^2 + a^2}$ ==
:$\ds \int \dfrac 1 {16 + x^2} \rd x = \dfrac 1 4 \arctan \dfrac x 4 + C$
== Proof ==
From Primitive of $\dfrac 1 {x^2 + a^2}$:
The result follows by setting $a = 4$.
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $13$.
Category:Examples of Use of Primitive of Reciprocal of x squared plus a squared
Primitive of Reciprocal of x squared plus a squared squared

== Theorem ==
:$\ds \int \frac {\d x} {\paren {x^2 + a^2}^2} = \frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac 1 {\paren {x^2 - a^2}^2}$
* Primitive of $\dfrac 1 {\paren {a^2 - x^2}^2}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.24$
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.132$
* : Front endpapers: A Brief Table of Integrals: $17$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.8.$
Category:Primitives involving x squared plus a squared
Primitive of Root of 2 a x minus x squared

== Theorem ==
:$\ds \int \sqrt {2 a x - x^2} \rd x = \frac {\paren {x - a} } 2 \sqrt {2 a x - x^2} + \frac {a^2} 2 \arcsin \frac {x - a} a + C$
where $C$ is an arbitrary constant.
== Proof ==
Let $u := x - a$.
Then:
:$\dfrac {\d u} {\d x} = 1$
and:
:$x = u + a$
Then:
and we have:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.48$
* : Front endpapers: A Brief Table of Integrals: $48$.
Category:Primitives of Roots of Quadratic Functions
Primitive of Root of 2 a x minus x squared over x

== Theorem ==
:$\ds \int \dfrac {\sqrt {2 a x - x^2} } x \rd x = \sqrt {2 a x - x^2} + a \arcsin \dfrac {x - a} a + C$
== Proof ==
Let $u := x - a$.
Then:
:$\dfrac {\d u} {\d x} = 1$
and:
:$x = u + a$
Then:
and we have:
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $52$.
Category:Primitives of Roots of Quadratic Functions
Primitive of Root of 2 a x minus x squared over x squared

== Theorem ==
:$\ds \int \dfrac {\sqrt {2 a x - x^2} } {x^2} \rd x = -2 \sqrt {\dfrac {2 a - x} x} - \arcsin \dfrac {x - a} a + C$
== 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:
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $53$.
Category:Primitives of Roots of Quadratic Functions