nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of Square of Hyperbolic Sine of a x by Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sinh^2 a x \cosh a x} = -\frac 1 a \map \arctan {\sinh a x} - \frac {\csch a x} a + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x $ and $\cosh a x$: $14.596$
Category:Primitives involving Hyperbolic Sine Function
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Square of Hyperbolic Sine of a x by Square of Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sinh^2 a x \cosh^2 a x} = \frac {-2 \coth 2 a x} a + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x $ and $\cosh a x$: $14.598$
* : $\S 17$: Tables of Special Indefinite Integrals: $(29)$ Integrals Involving $\sinh a x $ and $\cosh a x$: $17.29.5.$
Category:Primitives involving Hyperbolic Sine Function
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Square of Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sin^2 a x} = \frac {-\cot a x} a + C$
=== Corollary ===
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {\cos^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$: $14.351$
Category:Primitive of Reciprocal of Square of Sine of a x
Category:Primitives involving Sine Function
Primitive of Reciprocal of Square of Sine of a x/Corollary

== Corollary to Primitive of $\dfrac 1 {\sin^2 a x}$ ==
:$\ds \int \frac {\d x} {\sin^2 x} = -\cot x + C$
== Proof ==
== Sources ==
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
Category:Primitive of Reciprocal of Square of Sine of a x
Category:Primitives involving Sine Function
Primitive of Reciprocal of Square of Sine of a x by Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sin^2 a x \cos a x} = \frac 1 a \ln \size {\map \tan {\frac \pi 4 + \frac {a x} 2} } - \frac 1 {a \sin a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {\sin a x \cos^2 a x}$
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.405$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of Square of Sine of a x by Square of Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\sin^2 a x \cos^2 a x} = \frac {-2 \cot 2 a x} a + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sin a x$ and $\cos a x$: $14.407$
Category:Primitives involving Sine Function and Cosine Function
Primitive of Reciprocal of Square of p plus q by Exponential of a x

== Theorem ==
:$\ds \int \frac {\d x} {\paren {p + q e^{a x} }^2} = \frac x {p^2} + \frac 1 {a p \paren {p + q e^{a x} } } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $e^{a x}$: $14.516$
Category:Primitives involving Exponential Function
Primitive of Reciprocal of Square of p plus q by Hyperbolic Cosine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\paren {p + q \cosh a x}^2} = \frac {q \sinh a x} {a \paren {q^2 - p^2} \paren {p + q \cosh a x} } - \frac p {q^2 - p^2} \int \frac {\rd x} {p + q \cosh a x} + C$
== Proof ==
Hence the result.
== Also see ==
* Primitive of $\dfrac 1 {\paren {p + q \sinh a x}^2}$
== Sources ==
* : $\S 14$: Integrals involving $\cosh a x$: $14.582$
Category:Primitives involving Hyperbolic Cosine Function
Primitive of Reciprocal of Square of p plus q by Hyperbolic Sine of a x

== Theorem ==
:$\ds \int \frac {\d x} {\paren {p + q \sinh a x}^2} = \frac {-q \cosh a x} {a \paren {p^2 + q^2} \paren {p + q \sinh a x} } + \frac p {p^2 + q^2} \int \frac {\d x} {p + q \sinh a x} + C$
== Proof ==
Hence the result.
== Also see ==
* Primitive of $\dfrac 1 {\paren {p + q \cosh a x}^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sinh a x$: $14.554$
Category:Primitives involving Hyperbolic Sine Function
Primitive of Reciprocal of Tangent of a x

== Theorem ==
:$\ds \int \frac {\d x} {\tan a x} = \frac 1 a \ln \size {\sin a x} + C$
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {\sin a x}$
* Primitive of $\dfrac 1 {\cos a x}$
* Primitive of $\dfrac 1 {\cot a x}$
* Primitive of $\dfrac 1 {\sec a x}$
* Primitive of $\dfrac 1 {\csc a x}$
== Sources ==
* : $\S 14$: Integrals involving $\tan a x$: $14.434$
Category:Primitives involving Tangent Function
Primitive of Reciprocal of a squared minus x 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 $\dfrac 1 {x^2 - a^2}$
Category:Primitive of Reciprocal of a squared minus x squared
Category:Primitives involving a squared minus x squared
Category:Primitives involving Reciprocals
Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Cotangent Form

== Theorem ==
:$\ds \int \frac {\d x} {a^2 - x^2} = \frac 1 a \coth^{-1} \frac x a + C$
where $\size x > a$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {a^2 - x^2}$: $\tanh^{-1}$ form for the case $\size x < a$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Function Form

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

== Theorem ==
:$\ds \int \frac {\d x} {a^2 - x^2} = \frac 1 a \tanh^{-1} \frac x a + C$
where $\size x < a$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x^2 - a^2}$: $\coth^{-1}$ form for the case $\size x > a > 0$
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(vii)}$
* : $\S 14$: General Rules of Integration: $14.41$
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.163$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.41.$
* : $\S 17$: Tables of Special Indefinite Integrals: $(8)$ Integrals Involving $x^2 - a^2$, $x^2 < a^2$: $17.8.1.$
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x 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 a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form/Lemma

== Lemma ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Then:
:$\map \ln {\dfrac {a + x} {a - x} }$ is defined $\size x < a$
:$\map \ln {\dfrac {x + a} {x - a} }$ is defined $\size x > a$
== Proof ==
We have that the real natural logarithm is defined only on the strictly positive real numbers.
Hence:
:$\map \ln {\dfrac {a + x} {a - x} }$ is defined $\dfrac {a + x} {a - x} > 0$
:$\map \ln {\dfrac {x + a} {x - a} }$ is defined $\dfrac {x + a} {x - a} > 0$
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.
$\begin {array} {c|ccc|cc}
& x + a & a - x & x - a & \dfrac {a + x} {a - x} & \dfrac {x + a} {x - a} \\
\hline 
a < x & + & - & + & + & - \\
0 < x < a & + & + & - & - & + \\
-a < x < 0 & + & + & - & - & + \\
x < -a & - & + & - & + & - \\
\end {array}$
Hence:
:$\map \ln {\dfrac {a + x} {a - x} }$ is defined $-a < x < a$
:$\map \ln {\dfrac {x + a} {x - a} }$ is defined $x > a$ or $x < -a$
as we were required to show.
Category:Logarithms
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form/Partial Fraction Expansion

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

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$.
Then:
$\quad \ds \int \frac {\d x} {a^2 - x^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C & : \size x < a \\
& \\
\dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C & : \size x > a \\
& \\
\text {undefined} & : \size x = a \end {cases}$
== Proof ==
=== Case where $\size x < a$ ===
=== Case where $\size x > a$ ===
== Also see ==
* Primitive of $\dfrac 1 {x^2 + a^2}$
* Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x 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:
:$\ds \int \frac {\d x} {a^2 - x^2} = \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x greater than a/Proof 2

== Theorem ==
== Proof ==
Let $\size x > a$.
Then:
If $x > a$, then both $x + a > 0$ and $x - a > 0$.
So $\dfrac {x + a} {x - a} > 0$ and so:
:$\ln \size {\dfrac {x + a} {x - a} } = \map \ln {\dfrac {x + a} {x - a} }$
If $x < -a$, then both $x + a < 0$ and $x - a < 0$.
So again $\dfrac {x + a} {x - a} > 0$ and so:
:$\ln \size {\dfrac {x + a} {x - a} } = \map \ln {\dfrac {x + a} {x - a} }$
Hence the result.
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x 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:
:$\ds \int \frac {\d x} {a^2 - x^2} = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(vii)}$
* : $\S 14$: General Rules of Integration: $14.41$
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.163$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.41.$
* : $\S 17$: Tables of Special Indefinite Integrals: $(8)$ Integrals Involving $x^2 - a^2$, $x^2 < a^2$: $17.8.1.$
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x less than a/Proof 2

== Theorem ==
== Proof ==
Let $\size x < a$.
Then:
== Sources ==
* : $\text {II}$. Calculus: Integration: Algebraic Integration: $\text {I}$.
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form 2

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$ such that $\size x \ne a$.
:$\ds \int \frac {\d x} {a^2 - x^2} = \dfrac 1 {2 a} \ln \size {\dfrac {a + x} {a - x} } + C$
=== Corollary ===
== Proof ==
From the $1$st logarithm form:
:$\ds \int \frac {\d x} {a^2 - x^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C & : \size x < a\\ & \\ \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C & : \size x > a \\ & \\ \text {undefined} & : \size x = a \end {cases}$
From Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma:
:$\map \ln {\dfrac {a + x} {a - x} }$ is defined $\size x < a$
:$\map \ln {\dfrac {x + a} {x - a} }$ is defined $\size x > a$
Let $\size x < a$.
Then $\map \ln {\dfrac {a + x} {a - x} }$ is defined.
We have that:
So the result holds for $\size x < a$.
Let $\size x > a$.
Then $\map \ln {\dfrac {x + a} {x - a} }$ is defined.
We have:
We have that:
The result follows.
== Also presented as ==
== Sources ==
* : Chapter $\text {III}$: Integration: Integration
* : $\text {II}$. Calculus: Integration
* : Chapter $25$: Fundamental Integration Formulas: $22$.
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form 2/Also presented as

== Primitive of $\frac 1 {a^2 - x^2}$: Logarithm Form 2: Also presented as ==
This result is also seen presented in the form:
:$\ds \int \frac {\d x} {a^2 - x^2} = \dfrac 1 {2 a} \ln \size {\dfrac {x + a} {x - a} } + C$
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $18$.
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared/Logarithm Form 2/Corollary

== Theorem ==
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$ such that $\size x \ne a$.
:$\ds \int \frac {\d x} {a^2 - b^2 x^2} = \dfrac 1 {2 a b} \ln \size {\dfrac {a + b x} {a - b x} } + C$
== 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: $3.3.23$
Category:Primitive of Reciprocal of a squared minus x squared
Primitive of Reciprocal of a squared minus x squared squared

== Theorem ==
:$\ds \int \frac {\d x} {\paren {a^2 - x^2}^2} = \frac x {2 a^2 \paren {a^2 - x^2} } + \frac 1 {4 a^3} \map \ln {\frac {a + x} {a - x} } + C$
for $x^2 < a^2$.
== Proof ==
Hence the result.
== Also presented as ==
== Also see ==
* Primitive of $\dfrac 1 {\paren {x^2 - a^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.170$
Category:Primitive of Reciprocal of a squared minus x squared squared
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of a squared minus x squared squared/Also presented as

== Primitive of Reciprocal of a squared minus x squared squared: Also presented as ==
This result is also seen presented in the form:
:$\ds \int \frac {\d x} {\paren {a^2 - x^2}^2} = \frac x {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^2} \int \frac {\d x} {a^2 - x^2}$
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $19$.
Category:Primitive of Reciprocal of a squared minus x squared squared
Primitive of Reciprocal of a squared minus x squared squared/Partial Fraction Expansion

== Lemma for Primitive of Reciprocal of $\paren {a^2 - x^2}^2$ ==
:$\dfrac 1 {\paren {a^2 - x^2}^2} \equiv \dfrac 1 {4 a^2} \paren {\dfrac 1 {\paren {a - x}^2} + \dfrac 1 {\paren {a + x}^2} + \dfrac 1 {a \paren {a + x} } + \dfrac 1 {a \paren {a - x} } }$
== 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:Primitives involving a squared minus x squared/Lemmata
Primitive of Reciprocal of a x + b

== Theorem ==
:$\ds \int \frac {\d x} {a x + b} = \frac 1 a \ln \size {a x + b} + C$
where $a \ne 0$ and $x \ne - \dfrac b a$.
== Proof ==
== Examples ==
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $5$.
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(ii) (b)}$
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.15$
* : $\S 14$: Integrals involving $a x + b$: $14.59$
* : Front endpapers: A Brief Table of Integrals: $6$.
* : $9$ Sums and Integrals: $9.8$ Standard Integrals
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.1.$
Category:Primitive of Reciprocal of a x + b
Category:Primitives involving a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of a x + b/Examples/x - a

== Example of Use of Primitive of $\dfrac 1 {a x + b}$ ==
:$\ds \int \frac {\d x} {x - a} = \ln \size {x - a} + C$
== Proof ==
== Sources ==
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
Category:Examples of Use of Primitive of Reciprocal of a x + b
Primitive of Reciprocal of a x + b by p x + q

== Theorem ==
:$\ds \int \frac {\d x} {\paren {a x + b} \paren {p x + q} } = \frac 1 {b p - a q} \ln \size {\frac {p x + q} {a x + b} } + C$
where $b p \ne a q$.
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.20$
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.105$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.1.$
Category:Primitive of Reciprocal of a x + b by p x + q
Category:Primitives involving a x + b and p x + q
Primitive of Reciprocal of a x + b by p x + q/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\d x} {\paren {a x + b}^3} = \frac {-1} {2 a \paren {a x + b}^2} + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.73$
::(in which a mistake appears)
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.11.$
::(in which the same mistake appears)
Category:Primitive of Reciprocal of a x + b cubed
Category:Primitives involving a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of a x + b cubed/Mistake

== Source Work ==
:Chapter $14$: Indefinite Integrals:
::Integrals involving $a x + b$: $14.73$
:Chapter $17$: Tables of Special Indefinite Integrals:
::$(1)$ Integrals involving $a x + b$: $17.1.11.$
== Mistake ==
:$\ds \int \frac {\d x} {\paren {a x + b}^3} = \frac {-1} {2 \paren {a x + b}^2} + C$
== Correction ==
As demonstrated in Primitive of $\dfrac 1 {\paren {a x + b}^3}$ the correct expression is in fact:
:$\ds \int \frac {\d x} {\paren {a x + b}^3} = \frac {-1} {2 a \paren {a x + b}^2} + C$
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.73$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.11.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Primitive of Reciprocal of a x + b cubed/Proof 2

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

== Theorem ==
:$\ds \int \frac {\d x} {\paren {a x + b}^2} = -\frac 1 {a \paren {a x + b} } + C$
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.66$
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.6.$
Category:Primitive of Reciprocal of a x + b squared
Category:Primitives involving a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of a x + b squared/Examples/x - a squared

== Example of Use of Primitive of $\dfrac 1 {\paren {a x + b}^2}$ ==
:$\ds \int \frac {\d x} {\paren {x - a}^2} = -\frac 1 {\paren {x - a} } + C$
== Proof ==
== Sources ==
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
Category:Examples of Use of Primitive of Reciprocal of a x + b squared
Primitive of Reciprocal of a x + b squared/Proof 2

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

== Theorem ==
:$\ds \int \frac {\d x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac 1 {a x + b} + \frac p {b p - a q} \ln \size {\frac {p x + q} {a x + b} } } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.107$
* : $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.3.$
Category:Primitive of Reciprocal of a x + b squared by p x + q
Category:Primitives involving a x + b and p x + q
Primitive of Reciprocal of a x + b squared by p x + q/Partial Fraction Expansion

== Lemma for Primitive of $\frac 1 {\paren {a x + b}^2 \paren {p x + q} }$ ==
:$\dfrac 1 {\paren {a x + b}^2 \paren {p x + q} } \equiv \dfrac 1 {b p - a q} \paren {\dfrac {-a p} {\paren {b p - a q} \paren {a x + b} } + \dfrac {-a} {\paren {a x + b}^2} + \dfrac {p^2} {\paren {b p - a q} \paren {p x + q} } }$
== 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 Reciprocal of a x + b squared by p x + q
Category:Primitives involving a x + b and p x + q/Lemmata
Primitive of Reciprocal of a x squared + b by Root of c x squared + d

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers such that $a d \ne b c$.
Then:
:$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \begin {cases}
\dfrac 1 {\sqrt {b \paren {a d - b c} } } \arctan \dfrac {x \sqrt {a d - b c} } {\sqrt {b \paren {c x^2 + d} } } + C & : a d > b c \\
\dfrac 1 {2 \sqrt {b \paren {a d - b c} } } \ln \size {\dfrac {\sqrt {b \paren {c x^2 + d} } + x \sqrt {b c - a d} } {\sqrt {b \paren {c x^2 + d} } - x \sqrt {b c - a d} } } + C & : b c > a d
\end {cases}$
== Proof ==
=== Case $1$: $a d > b c$ ===
=== Case $2$: $b c > a d$ ===
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions
Category:Primitive of Reciprocal of a x squared + b by Root of c x squared + d
Category:Primitives of Roots of Quadratic Functions
Primitive of Reciprocal of a x squared + b by Root of c x squared + d/Lemma

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers such that $a d \ne b c$.
Then:
:$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac {\sqrt c} {2 a} \int \dfrac {\d u} {\paren {\sqrt {\paren {u - \frac d 2}^2 - \frac d 4} } \paren {u - \frac {a d + b c} a} }$
where $u := c x^2 + d$.
== Proof ==
Hence:
Category:Primitive of Reciprocal of a x squared + b by Root of c x squared + d
Primitive of Reciprocal of a x squared + b by Root of c x squared + d/a d greater than b c

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers such that $a d \ne b c$.
Let $a d > b c$.
Then:
:$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac 1 {\sqrt {b \paren {a d - b c} } } \arctan \dfrac {x \sqrt {a d - b c} } {\sqrt {b \paren {c x^2 + d} } } + C$
== Proof ==
=== Lemma ===
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.49$
Category:Primitive of Reciprocal of a x squared + b by Root of c x squared + d