nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form

== Theorem ==
:$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$
where $a$ is a strictly positive constant and $a^2 > x^2$.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
Substitute:
:$\sin \theta = \dfrac x a \iff x = a \sin \theta$
for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Real Sine Function is Bounded and Shape of Sine Function, this substitution is valid for all $x / a \in \openint {-1} 1$.
Then:
so this substitution will not change the domain of the integrand.
Then:
We have defined $\theta$ to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Sine and Cosine are Periodic on Reals, $\cos \theta > 0$ for the entire interval.
Therefore the absolute value is unnecessary, and the integral simplifies to:
As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
:$\sin \theta = \dfrac x a \iff \theta = \arcsin \dfrac x a$
The answer in terms of $x$, then, is:
:$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$
== Also see ==
* Primitive of Reciprocal of $\sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $\sqrt {x^2 - a^2}$
* Derivative of Arcsine Function
== Sources ==
* : Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $12$.
* : Chapter $\text {III}$: Integration: Integration
* : $\text {II}$. Calculus: Integration: Algebraic Integration: $\text {III}$.
* : $\text {II}$. Calculus: Integration
* : Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(ix)}$
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.44$
* : $\S 14$: General Rules of Integration: $14.42$
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.237$
* : Front endpapers: A Brief Table of Integrals: $28$.
* : Chapter $25$: Fundamental Integration Formulas: $18$.
* : Chapter $5$. Integrals: Integrals of Special Functions: $25$
* : $6$. 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: $16.42.$
* : $\S 17$: Tables of Special Indefinite Integrals: $(11)$ Integrals Involving $\sqrt {a^2 - x^2}$: $17.11.1.$
* : Appendix $7$: Integrals
* : Appendix $8$: Integrals
* 
Category:Primitive of Reciprocal of Root of a squared minus x squared
Category:Arcsine Function
Primitive of Reciprocal of Root of a squared minus x squared/Examples/16 - x^2

== Example of Use of Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ ==
:$\ds \int \dfrac 1 {\sqrt {16 - x^2} } \rd x = \arcsin \dfrac x 4 + C$
== Proof ==
From Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$:
The result follows by setting $a = 4$.
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $12$.
Category:Examples of Use of Primitive of Reciprocal of Root of a squared minus x squared
Primitive of Reciprocal of Root of a squared minus x squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {\paren {\sqrt {a^2 - x^2} }^3} = \frac x {a^2 \sqrt {a^2 - x^2} } + C$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac 1 {\paren {\sqrt {x^2 - a^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.251$
Category:Primitive of Reciprocal of Root of a squared minus x squared cubed
Category:Primitives involving Root of a squared minus x squared
Primitive of Reciprocal of Root of a x + b

== Theorem ==
:$\ds \int \frac {\d x} {\sqrt{a x + b} } = \frac {2 \sqrt {a x + b} } a + C$
where $a x + b > 0$.
== Proof ==
Put $u = \sqrt{a x + b}$.
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.84$
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.1.$
Category:Primitive of Reciprocal of Root of a x + b
Category:Primitives involving Root of a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of Root of a x + b by Root of p x + q

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
Then:
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C & : a p < 0 \\ \\
\dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C & : a > 0, p < 0 \\ \\
\ds \dfrac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C & : a p > 0
\end {cases}$
for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.
=== Case where $a p = b q$ ===
== Proof ==
=== Case $1$: $a p < 0$ ===
=== Case $2$: $a > 0, p < 0$ ===
=== Case $3$: $a p > 0$ ===
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $14.120$
* : $\S 17$: Tables of Special Indefinite Integrals: $(5)$ Integrals Involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $17.5.1.$
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Category:Primitives involving Root of a x + b and Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/Completing the Square

== Lemma for Primitive of $\frac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$ ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
Then:
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\ds \frac 1 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {b p - a q}^2} } & : a p > 0 \\
\ds \frac 1 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {b p - a q}^2 - u^2} } & : a p < 0
\end {cases}$
where:
:$u := 2 a p x + b p + a q$
== Proof ==
Let us make the substitution:
;Case $1: \quad a p > 0$
;Case $2: \quad a p < 0$
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/Lemma 1

== Lemma for Primitive of $\frac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$ ==
Let $u = \sqrt {a x + b}$.
Then:
:$\ds \sqrt {p x + q} = \sqrt {\paren {\frac p a} \paren {u^2 - \paren {\frac {b p - a q} p} } }$
== Proof ==
We have:
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/Lemma 2

== Lemma for Primitive of $\frac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$ ==
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\ds \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } } & : a p > 0 \\
\ds \frac 2 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {\frac {b p - a q} p} - u^2} } & : a p < 0
\end {cases}$
where:
:$u := \sqrt {a x + b}$
== Proof ==
Let us make the substitution:
:$u = \sqrt {a x + b}$
=== Lemma ===
;Case $1: \quad a p > 0$
;Case $2: \quad a p < 0$
We have:
Then:
Then:
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/Mistake

== Source Work ==
:Chapter $14$: Indefinite Integrals
::Integrals involving $\sqrt {a x + b}$ and $\sqrt{p x + q}$: $14.120$
:Chapter $17$: Tables of Special Indefinite Integrals
::$(5)$ Integrals Involving $\sqrt {a x + b}$ and $\sqrt{p x + q}$: $17.5.1.$
== Mistake ==
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\dfrac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } \\ \\
\dfrac 2 {\sqrt {-a p} } \tan^{-1} \sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } }
\end{cases}$
== Correction ==
It is important to give the conditions under which the primitives are valid.
It is also important to specify that the argument of the logarithm function is absolute:
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
\dfrac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } & : a p > 0 \\ \\
\dfrac 2 {\sqrt {-a p} } \tan^{-1} \sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } & : a p < 0 
\end{cases}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $14.120$
* : $\S 17$: Tables of Special Indefinite Integrals: $(5)$ Integrals Involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $17.5.1.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Primitive of Reciprocal of Root of a x + b by Root of p x + q/a greater than 0, p less than 0

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
Let $a > 0$ and $p < 0$.
Then:
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C$
for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.
== Proof ==
=== Completing the Square ===
Hence:
</onlyinclude>
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.27$
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p greater than 0

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
Let $a p > 0$.
Then:
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C$
for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.
== Proof ==
=== Lemma $1$ ===
=== Lemma $2$ ===
We have that $a p > 0$.
Thus:
Let $\dfrac {b p - a q} p > 0$.
Then we set:
:$(2): \quad c^2 :=\dfrac {b p - a q} p$
Then:
Let $\dfrac {b p - a q} p < 0$.
Then we set:
:$(3): \quad c^2 := -\dfrac {b p - a q} p$
Then:
Thus we have in both cases that:
$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \frac 2 {\sqrt {a p} } \ln \size {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C$
Then:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.28$
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p greater than 0/Mistake

== Source Work ==
:$3$: Elementary Analytic Methods
::$3.3$ Rules for Differentiation and Integration:
:::Integrals of Irrational Algebraic Functions: $3.3.28$
== Mistake ==
:$\ds \int \frac {\d x} {\sqbrk {\paren {a + b x} \paren {c + d x} }^{1/2} } = \dfrac 2 {\paren {b d}^{1/2} } \ln \size {\sqbrk {b d \paren {a + b x} }^{1/2} + \sqbrk {b \paren {c + d x} }^{1/2} } + C$
== Correction ==
This should read:
:$\ds \int \frac {\d x} {\sqbrk {\paren {a + b x} \paren {c + d x} }^{1/2} } = \dfrac 2 {\paren {b d}^{1/2} } \ln \size {\sqbrk {d \paren {a + b x} }^{1/2} + \sqbrk {b \paren {c + d x} }^{1/2} } + C$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.28$
Category:Mistakes/Handbook of Mathematical Functions
Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
Let $a p < 0$.
Then:
:$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C$
for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.26$
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0/Proof 1

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
== Proof ==
Let us make the substitution:
Then:
Hence:
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x + b by Root of p x + q/a p less than 0/Proof 2

== Theorem ==
Let $a, b, p, q \in \R$ such that $a p \ne b q$.
== Proof ==
First let us express the integrand in the following form:
Recall the definition of Euler's third substitution:
In this context we have:
Hence we make the substitution:
Then we use:
Then:
Then:
Assembling the pieces:
== Sources ==
* 
Category:Primitive of Reciprocal of Root of a x + b by Root of p x + q
Primitive of Reciprocal of Root of a x squared plus b x plus c

== Theorem ==
=== $a$ greater than $0$ ===
=== $a$ less than $0$ ===
=== $a$ equal to Zero ===
When $a = 0$ we have the following result:
== Examples ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.280$
* : $\S 17$: Tables of Special Indefinite Integrals: $(13)$ Integrals Involving $\sqrt {a x^2 + b x + c}$: $17.13.1.$
Category:Primitives involving Root of a x squared plus b x plus c
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples

== Examples of Use of Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ ==
=== Primitive of $\dfrac 1 {\sqrt {2 + 4 x - 3 x^2} }$ ===
=== Primitive of $\dfrac 1 {\sqrt {x^2 + 4 x + 5} }$ ===
=== Primitive of $\dfrac 1 {\sqrt {2 x - x^2} }$ ===
=== Primitive of $\dfrac 1 {\sqrt {2 x + x^2} }$ ===
Category:Examples of Use of Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 + 4 x - 3 x^2

== Example of Use of Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ ==
:$\ds \int \dfrac {\d x} {\sqrt {2 + 4 x - 3 x^2} } = \dfrac 1 {\sqrt 3} \map \arcsin {\dfrac {3 x - 2} {10} } + C$
== Proof 1 ==
== Proof 2 ==
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 + 4 x - 3 x^2
Category:Examples of Use of Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 + 4 x - 3 x^2/Proof 1

== Example of Use of Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ ==
== Proof ==
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 + 4 x - 3 x^2
Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 + 4 x - 3 x^2/Proof 2

== Example of Use of Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ ==
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Integration: Algebraic Integration: Example
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 + 4 x - 3 x^2
Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 x + x^2

== Example of Use of Primitive of $\dfrac 1 {a x^2 + b x + c}$ ==
:$\ds \int \dfrac {\d x} {2 x + x^2} = $
== Proof ==
We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:
We note that:
Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C$
Substituting for $a$, $b$ and $c$ and simplifying:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $20$.
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/2 x - x^2

== Example of Use of Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ ==
:$\ds \int \dfrac {\d x} {\sqrt {2 x - x^2} } = \map \arcsin {x - 1} + C$
== Proof ==
From Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ for negative $a$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac {-1} {\sqrt {-a} } \map \arcsin {\dfrac {2 a x + b} {\sqrt {\size {b^2 - 4 a c} } } } + C$
as long as $b^2 - 4 a c \ne 0$.
Substituting for $a$, $b$ and $c$ and simplifying:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $19$.
Category:Examples of Use of Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/Examples/x^2 + 4 x + 5

== Example of Use of Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ ==
:$\ds \int \dfrac {\d x} {\sqrt {x^2 + 4 x + 5} } = \map \ln {x + 2 + \sqrt {x^2 + 4 x + 5} } + C$
== Proof ==
We aim to use Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$ with:
We note that:
Hence from Primitive of $\dfrac 1 {\sqrt {a x^2 + b x + c} }$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac 1 {\sqrt a} \map \arsinh {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
Substituting for $a$, $b$ and $c$ and simplifying:
== Sources ==
* : $\text {II}$. Calculus: Exercises $\text {XIV}$: $16$.
Category:Examples of Use of Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a equal to 0

== Theorem ==
For $a x^2 + b x + c > 0$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {2 \sqrt {b x + c} } b + C$
when $a = 0$.
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.280$
* : $\S 17$: Tables of Special Indefinite Integrals: $(13)$ Integrals Involving $\sqrt {a x^2 + b x + c}$: $17.13.1.$
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a greater than 0

== Theorem ==
Let $a \in \R_{> 0}$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \begin {cases}
\dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C & : b^2 - 4 a c > 0 \\ \\
\dfrac 1 {\sqrt a} \map \arsinh {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt a} \ln \size {2 a x + b} + C & : b^2 - 4 a c = 0
\end {cases}$
where $\arsinh$ denotes the area hyperbolic sine function.
== Proof ==
=== Completing the Square ===
=== Positive Discriminant ===
=== Negative Discriminant ===
=== Zero Discriminant ===
== Sources ==
* : $\text {II}$. Calculus: Integration: Algebraic Integration: Type $\text D$.
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a greater than 0/Negative Discriminant

== Theorem ==
Let $a \in \R_{>0}$.
Let $b^2 - 4 a c < 0$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac 1 {\sqrt a} \map \arsinh {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
where $\arsinh$ denotes the area hyperbolic sine function.
== Proof ==
=== Completing the Square ===
Let $b^2 - 4 a c < 0$.
Then:
Thus:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.34$
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a greater than 0/Positive Discriminant

== Theorem ==
Let $a, b, c \in \R$ such that $a > 0$.
Let $b^2 - 4 a c > 0$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C$
== Proof ==
=== Completing the Square ===
Let $b^2 - 4 a c > 0$.
Then:
Thus:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.33$
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a greater than 0/Zero Discriminant

== Theorem ==
Let $a \in \R_{>0}$.
Let $b^2 - 4 a c = 0$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac 1 {\sqrt a} \ln \size {2 a x + b} + C$
== Proof ==
Let $b^2 - 4 a c = 0$ .
Then:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.35$
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.280$
* : $\S 17$: Tables of Special Indefinite Integrals: $(13)$ Integrals Involving $\sqrt {a x^2 + b x + c}$: $17.13.1.$
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a less than 0

== Theorem ==
Let $a \in \R_{< 0}$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac {-1} {\sqrt {-a} } \map \arcsin {\dfrac {2 a x + b} {\sqrt {\size {b^2 - 4 a c} } } } + C$
given that $b^2 \ne 4 a c$.
== Proof ==
=== Completing the Square ===
=== Positive Discriminant ===
=== Negative Discriminant ===
=== Zero Discriminant ===
In summary:
:For $b^2 - 4 a c > 0: \ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {-1} {\sqrt {-a} } \map \arcsin {\frac {2 a x + b} {\sqrt {b^2 - 4 a c} } } + C$
:For $b^2 - 4 a c < 0: \ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {-1} {\sqrt {-a} } \map \arcsin {\frac {2 a x + b} {\sqrt {-\paren {b^2 - 4 a c} } } } + C$
and so by definition of absolute value:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac {-1} {\sqrt {-a} } \map \arcsin {\dfrac {2 a x + b} {\sqrt {\size {b^2 - 4 a c} } } } + C$
== Sources ==
* : $\text {II}$. Calculus: Integration: Algebraic Integration: Type $\text D$.
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a less than 0/Negative Discriminant

== Theorem ==
Let $a \in \R_{> 0}$.
Let $b^2 - 4 a c > 0$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {-1} {\sqrt {-a} } \map \arcsin {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
== Proof ==
=== Completing the Square ===
Let $b^2 - 4 a c < 0$.
Let $D' = -D = 4 a c - b^2$.
Then:
Thus:
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a less than 0/Positive Discriminant

== Theorem ==
Let $a \in \R_{<0}$.
Let $b^2 - 4 a c > 0$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$ and $\size {2 a x + b} < \sqrt {b^2 - 4 a c}$:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {-1} {\sqrt {-a} } \map \arcsin {\frac {2 a x + b} {\sqrt {b^2 - 4 a c} } } + C$
== Proof ==
=== Completing the Square ===
Let $b^2 - 4 a c > 0$.
Then:
Thus:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.36$
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of a x squared plus b x plus c/a less than 0/Zero Discriminant

== Theorem ==
Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c = 0$.
Then:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
is not defined.
== Proof ==
Suppose that $b^2 - 4 a c = 0$.
Then:
But we have that:
:$\paren {2 a x + b}^2 > 0$
while under our assertion that $a < 0$:
:$4 a < 0$
and so:
:$a x^2 + b x + c < 0$
Thus on the real numbers $\sqrt {a x^2 + b x + c}$ is not defined.
Hence it follows that:
:$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
is not defined.
Category:Primitive of Reciprocal of Root of a x squared plus b x plus c
Primitive of Reciprocal of Root of x squared minus a squared

== Theorem ==
=== Inverse Hyperbolic Cosine Form ===
=== Logarithm Form ===
== Also see ==
* Primitive of Reciprocal of $\sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $\sqrt {a^2 - x^2}$
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $36$.
Category:Primitive of Reciprocal of Root of x squared minus a squared
Category:Primitives involving Root of x squared minus a squared
Category:Primitives involving Reciprocals