nyuuzyou/wikis huggingface dataset:
Primitive of Root of Function under Half its Derivative

== Theorem ==
Let $f$ be a real function which is integrable.
Then:
:$\ds \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x = \sqrt {\map f x} + C$
where $C$ is an arbitrary constant.
== Proof ==
By Integration by Substitution (with appropriate renaming of variables):
:$\ds \int \map g u \rd u = \int \map g {\map f x} \map {f'} x \rd x$
Let $\map u x = \sqrt {\map f x}$
Then:
== Sources ==
* : Chapter $10$: Integration: $10.4$. Standard integrals: General Rules: $\text {IV}$.
Category:Primitives
Primitive of Root of a squared minus x squared/Arccosine Form

== Theorem ==
:$\ds \int \sqrt {a^2 - x^2} \rd x = \frac {x \sqrt {a^2 - x^2} } 2 - \frac {a^2} 2 \arccos \frac x a + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\sqrt {x^2 + a^2}$
* Primitive of $\sqrt {x^2 - a^2}$
== Sources ==
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitive of Root of a squared minus x squared
Primitive of Root of a squared minus x squared/Arcsine Form

== Theorem ==
:$\ds \int \sqrt {a^2 - x^2} \rd x = \frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C$
== Proof ==
Let:
Also:
and:
Thus:
== Also see ==
* Primitive of $\sqrt {x^2 + a^2}$
* Primitive of $\sqrt {x^2 - a^2}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.45$
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.244$
* : Front endpapers: A Brief Table of Integrals: $29$.
* : Chapter $25$: Fundamental Integration Formulas: $25$.
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
* : Appendix $7$: Integrals
* : Appendix $8$: Integrals
Category:Primitive of Root of a squared minus x squared
Primitive of Root of a squared minus x squared/Arcsine Form/Mistake

== Source Work ==
:
:Appendix $7$: Integrals
== Mistake ==
:$\ds \int \sqrt {x^2 + x^2} \rd x = \frac 1 2 x \sqrt {x^2 + x^2} + \frac 1 2 a^2 \sin^{-1} \frac x a$
== Correction ==
This should read:
:$\ds \int \sqrt {a^2 - x^2} \rd x = \frac 1 2 x \sqrt {a^2 - x^2} + \frac 1 2 a^2 \sin^{-1} \frac x a$
This has been corrected in the $6$th edition.
== Sources ==
* : Appendix $7$: Integrals
Category:Mistakes/The Concise Oxford Dictionary of Mathematics
Primitive of Root of a squared minus x squared cubed

== Theorem ==
:$\ds \int \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\paren {\sqrt {x^2 + a^2} }^3$
* Primitive of $\paren {\sqrt {x^2 - a^2} }^3$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.258$
Category:Primitives involving Root of a squared minus x squared
Primitive of Root of a squared minus x squared cubed over x

== Theorem ==
:$\ds \int \frac {\paren {\sqrt {a^2 - x^2} }^3} x \rd x = \frac {\paren {\sqrt {a^2 - x^2} }^3} 3 + a^2 \sqrt {a^2 - x^2} - a^3 \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\paren {\sqrt {x^2 + a^2} }^3} x$
* Primitive of $\dfrac {\paren {\sqrt {x^2 - a^2} }^3} x$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.262$
Category:Primitives involving Root of a squared minus x squared
Primitive of Root of a squared minus x squared cubed over x cubed

== Theorem ==
:$\ds \int \frac {\paren {\sqrt {a^2 - x^2} }^3} {x^3} \rd x = \frac {-\paren {\sqrt {a^2 - x^2} }^3} {2 x^2} - \frac {3 \sqrt {a^2 - x^2} } 2 + \frac {3 a} 2 \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\paren {\sqrt {x^2 + a^2} }^3} {x^3}$
* Primitive of $\dfrac {\paren {\sqrt {x^2 - a^2} }^3} {x^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.264$
* : $\S 17$: Tables of Special Indefinite Integrals: $(11)$ Integrals Involving $\sqrt {a^2 - x^2}$: $17.11.28.$
Category:Primitives involving Root of a squared minus x squared
Primitive of Root of a squared minus x squared cubed over x squared

== Theorem ==
:$\ds \int \frac {\paren {\sqrt {a^2 - x^2} }^3} {x^2} \rd x = \frac {-\paren {\sqrt {a^2 - x^2} }^3} x - \frac {3 x \sqrt {a^2 - x^2} } 2 - \frac {3 a^2} 2 \arcsin \frac x a + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\paren {\sqrt {x^2 + a^2} }^3} {x^2}$
* Primitive of $\dfrac {\paren {\sqrt {x^2 - a^2} }^3} {x^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.263$
Category:Primitives involving Root of a squared minus x squared
Primitive of Root of a squared minus x squared over x

== Theorem ==
:$\ds \int \frac {\sqrt {a^2 - x^2} } x \rd x = \sqrt {a^2 - x^2} - a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$
== Proof ==
Let:
== Also presented as ==
== Also see ==
* Primitive of $\dfrac {\sqrt {x^2 + a^2} } x$
* Primitive of $\dfrac {\sqrt {x^2 - a^2} } x$
* Cartesian Equation of Tractrix
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.248$
Category:Primitive of Root of a squared minus x squared over x
Category:Primitives involving Root of a squared minus x squared
Primitive of Root of a squared minus x squared over x/Also presented as

== Primitive of $\frac {\sqrt {a^2 - x^2} } x$: Also presented as ==
This result is also seen presented in the form:
:$\ds \int \frac {\sqrt {a^2 - x^2} } x \rd x = \sqrt {a^2 - x^2} - a \ln \size {\frac {a + \sqrt {a^2 - x^2} } x} + C$
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $31$.
Category:Primitive of Root of a squared minus x squared over x
Primitive of Root of a squared minus x squared over x cubed

== Theorem ==
:$\ds \int \frac {\sqrt {a^2 - x^2} } {x^3} \rd x = \frac {-\sqrt {a^2 - x^2} } {2 x^2} + \frac 1 {2 a} \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\sqrt {x^2 + a^2} } {x^3}$
* Primitive of $\dfrac {\sqrt {x^2 - a^2} } {x^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.250$
Category:Primitives involving Root of a squared minus x squared
Primitive of Root of a squared minus x squared over x squared

== Theorem ==
:$\ds \int \frac {\sqrt {a^2 - x^2} } {x^2} \rd x = \frac {-\sqrt {a^2 - x^2} } x - \arcsin \frac x a + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\sqrt {x^2 + a^2} } {x^2}$
* Primitive of $\dfrac {\sqrt {x^2 - a^2} } {x^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.249$
* : Front endpapers: A Brief Table of Integrals: $32$.
Category:Primitives involving Root of a squared minus x squared
Primitive of Root of a x + b

== Theorem ==
:$\ds \int \sqrt {a x + b} \rd x = \frac {2 \sqrt {\paren {a x + b}^3} } {3 a}$
== Proof ==
Let $u = \sqrt{a x + b}$.
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.89$
* : $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.6.$
Category:Primitives involving Root of a x + b
Primitive of Root of a x + b by Root of p x + q

== Theorem ==
:$\ds \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
== Proof ==
From Primitive of $\paren {p x + q}^n \sqrt {a x + b}$:
:$\ds \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x$
Putting $n = \dfrac 1 2$:
== Also presented as ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.31$
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $14.122$
* : $\S 17$: Tables of Special Indefinite Integrals: $(5)$ Integrals Involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $17.5.3.$
Category:Primitive 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 Root of a x + b by Root of p x + q/Also presented as

== Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$: Also presented as ==
This result can also be seen presented in this form:
:$\ds \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x = \frac {\paren {b p - a q} + 2 a {p x + q} } {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.31$
Category:Primitive of Root of a x + b by Root of p x + q
Primitive of Root of a x + b over Power of p x + q

== Theorem ==
:$\ds \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x = \frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$
== Proof ==
From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 3:
:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m a \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^{n - 1} } \rd x}$
Setting $m := \dfrac 1 2$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.119$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.7.$
Category:Primitives involving Root of a x + b and p x + q
Primitive of Root of a x + b over Power of x/Formulation 1

== Theorem ==
:$\ds \int \frac {\sqrt {a x + b} } {x^m} \rd x = -\frac {\sqrt {a x + b} } {\paren {m - 1} x^{m - 1} } + \frac a {2 \paren {m - 1} } \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$
== Proof ==
Let:
From Integration by Parts:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
from which:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.97$
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.14.$
Category:Primitive of Root of a x + b over Power of x
Primitive of Root of a x + b over Power of x/Formulation 2

== Theorem ==
:$\ds \int \frac {\sqrt{a x + b} } {x^m} \rd x = -\frac {\paren {\sqrt{a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 5} a} {\paren {2 m - 2} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x$
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Putting $n := \dfrac 1 2$ and $m := -m$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.98$
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.15.$
Category:Primitive of Root of a x + b over Power of x
Primitive of Root of a x + b over p x + q

== Theorem ==
:$\ds \int \frac {\sqrt{a x + b} } {p x + q} \rd x = \begin{cases}
\dfrac {2 \sqrt{a x + b} } p + \dfrac {\sqrt {b p - a q} } {p \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \\
\dfrac {2 \sqrt{a x + b} } p - \dfrac {\sqrt {a q - b p} } {p \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\
\end{cases}$
== Proof ==
From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 2:
:$\ds \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$
Setting $m := \dfrac 1 2$ and $n = 1$:
From Primitive of Reciprocal of $p x + q$ by Root of $a x + b$:
:$\ds \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} } = \begin{cases}
\dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \\
\dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\
\end{cases}$
The result follows by substitution.
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.115$
* : $\S 17$: Tables of Special Indefinite Integrals: $(4)$ Integrals Involving $\sqrt {a x + b}$ and $p x + q$: $17.4.3.$
Category:Primitives involving Root of a x + b and p x + q
Primitive of Root of a x + b over x

== Theorem ==
:$\ds \int \frac {\sqrt {a x + b} } x \rd x = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Primitive of Reciprocal of $x \sqrt {a x + b}$ for $\ds \int \frac {\d x} {x \sqrt {a x + b} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.92$
* : Front endpapers: A Brief Table of Integrals: $12$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.9.$
Category:Primitive of Root of a x + b over x
Category:Primitives involving Root of a x + b
Primitive of Root of a x + b over x/Proof 1

== Theorem ==
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
Putting $m = -1$ and $n = \dfrac 1 2$:
Category:Primitive of Root of a x + b over x
Primitive of Root of a x + b over x/Proof 2

== Theorem ==
== Proof ==
Let:
From Integration by Parts:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \, \frac {\d u} {\d x} \rd x$
from which:
Category:Primitive of Root of a x + b over x
Primitive of Root of a x + b over x squared

== Theorem ==
:$\ds \int \frac {\sqrt {a x + b} } {x^2} \rd x = -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }$
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
Putting $m = -2$ and $n = \dfrac 1 2$:
== Also see ==
* Primitive of Reciprocal of $x \sqrt {a x + b}$ for $\ds \int \frac {\d x} {x \sqrt {a x + b} }$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.93$
* : Front endpapers: A Brief Table of Integrals: $14$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.10.$
Category:Primitives involving Root of a x + b
Primitive of Root of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \sqrt {a x^2 + b x + c} \rd x = \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$
== Proof ==
Let:
Suppose $a > 0$.
Then we have:
When $a < 0$, the above does not work, as we cannot take the square root of a negative number.
Hence in this case:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.37$
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.285$
Category:Primitives involving Root of a x squared plus b x plus c
Primitive of Root of a x squared plus b x plus c over x

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\sqrt {a x^2 + b x + c} } x \rd x = \sqrt {a x^2 + b x + c} + \frac b 2 \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + bx + c}$: $14.288$
Category:Primitives involving Root of a x squared plus b x plus c
Primitive of Root of a x squared plus b x plus c over x squared

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\sqrt {a x^2 + b x + c} } {x^2} \rd x = \frac {-\sqrt {a x^2 + b x + c} } x + a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + \frac b 2 \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }$ 
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + bx + c}$: $14.289$
Category:Primitives involving Root of a x squared plus b x plus c
Primitive of Root of p x + q over Root of a x + b

== Theorem ==
:$\ds \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
== Proof ==
From Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$:
:$\ds \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$
Putting $n = \dfrac 1 2$:
== Also presented as ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.32$
* : $\S 14$: Integrals involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $14.123$
* : $\S 17$: Tables of Special Indefinite Integrals: $(5)$ Integrals Involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$: $17.5.4.$
Category:Primitive of Root of p x + q over Root of a x + b
Category:Primitives involving Root of a x + b and Root of p x + q
Primitive of Root of p x + q over Root of a x + b/Also presented as

== Primitive of $\sqrt {\frac {p x + q} {a x + b} }$: Also presented as ==
This result can also be seen presented in this form:
$\ds \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a - \frac {b p - a q} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.32$
Category:Primitive of Root of p x + q over Root of a x + b
Primitive of Root of x squared minus a squared

== Theorem ==
=== Inverse Hyperbolic Cosine Form ===
=== Logarithm Form ===
== Also see ==
* Primitive of $\sqrt{x^2 + a^2}$
* Primitive of $\sqrt{a^2 - x^2}$
Category:Primitive of Root of x squared minus a squared
Category:Primitives involving Root of x squared minus a squared
Primitive of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form

== Theorem ==
:$\ds \int \sqrt {x^2 - a^2} \rd x = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \cosh^{-1} \frac x a + C$
== Proof ==
Let:
Also:
and:
Thus:
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $37$.
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitive of Root of x squared minus a squared
Primitive of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form/Mistake

== Source Work ==
:Table 2 Integrals
== Mistake ==
:$\ds \int \sqrt {x^2 - a^2} \rd x = \frac 1 2 a^2 \cosh^{-1} \frac x a + \frac 1 2 {x \sqrt {x^2 - a^2} } + C$
== Correction ==
Should be:
:$\ds \int \sqrt {x^2 - a^2} \rd x = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \cosh^{-1} \frac x a + C$
== Sources ==
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Mistakes/The Penguin Dictionary of Mathematics
Primitive of Root of x squared minus a squared/Logarithm Form

== Theorem ==
:$\ds \int \sqrt {x^2 - a^2} \rd x = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C$
for $\size x \ge a$.
== Proof ==
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 \ge a^2$, that is, either:
:$x \ge a$
or:
:$x \le -a$
where it is assumed that $a > 0$.
First let $x \ge a$.
Also:
and:
Thus:
Now suppose $x \le -a$.
Let $z = -x$.
Then:
:$\d x = -\d z$
and we then have:
The result follows.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.41$
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.216$
* : Chapter $25$: Fundamental Integration Formulas: $27$.
* : Appendix $7$: Integrals
* : Appendix $8$: Integrals
Category:Primitive of Root of x squared minus a squared
Primitive of Root of x squared minus a squared cubed

== Theorem ==
:$\ds \int \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C$
for $\size x \ge a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\paren {\sqrt {x^2 + a^2} }^3$
* Primitive of $\paren {\sqrt {a^2 - z^2} }^3$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.230$
Category:Primitives involving Root of x squared minus a squared
Primitive of Root of x squared minus a squared cubed over x

== Theorem ==
:$\ds \int \frac {\paren {\sqrt {x^2 - a^2} }^3} x \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 - a^2 \sqrt {x^2 - a^2} + a^3 \arcsec \size {\frac x a} + C$
for $\size x \ge a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\paren {\sqrt {x^2 + a^2} }^3} x$
* Primitive of $\dfrac {\paren {\sqrt {a^2 - x^2} }^3} x$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.234$
Category:Primitives involving Root of x squared minus a squared
Primitive of Root of x squared minus a squared cubed over x cubed

== Theorem ==
:$\ds \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^3} \rd x = \frac {-\paren {\sqrt {x^2 - a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 - a^2} } 2 - \frac {3 a} 2 \arcsec \size {\frac x a} + C$
for $\size x \ge a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\paren {\sqrt {x^2 + a^2} }^3} {x^3}$
* Primitive of $\dfrac {\paren {\sqrt {a^2 - x^2} }^3} {x^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.236$
* : $\S 17$: Tables of Special Indefinite Integrals: $(10)$ Integrals Involving $\sqrt {x^2 - a^2}$: $17.10.27.$
Category:Primitives involving Root of x squared minus a squared
Primitive of Root of x squared minus a squared cubed over x squared

== Theorem ==
:$\ds \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^2} \rd x = \frac {-\paren {\sqrt {x^2 - a^2} }^3} x + \frac{3 x \sqrt {x^2 - a^2} } 2 - \frac {3 a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C$
for $\size x \ge a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\paren {\sqrt {x^2 + a^2} }^3} {x^2}$
* Primitive of $\dfrac {\paren {\sqrt {a^2 - x^2} }^3} {x^2}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.235$
Category:Primitives involving Root of x squared minus a squared
Primitive of Root of x squared minus a squared over x

== Theorem ==
:$\ds \int \frac {\sqrt {x^2 - a^2} } x \rd x = \sqrt {x^2 - a^2} - a \arcsec \size {\frac x a} + C$
for $\size x \ge a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\sqrt {x^2 + a^2} } x$
* Primitive of $\dfrac {\sqrt {a^2 - x^2} } x$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.220$
* : Front endpapers: A Brief Table of Integrals: $42$.
Category:Primitives involving Root of x squared minus a squared
Primitive of Root of x squared minus a squared over x cubed

== Theorem ==
:$\ds \int \frac {\sqrt {x^2 - a^2} } {x^3} \rd x = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 {2 a} \arcsec \size {\frac x a} + C$
for $\size x \ge a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac {\sqrt {x^2 + a^2} } {x^3}$
* Primitive of $\dfrac {\sqrt {a^2 - x^2} } {x^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.222$
Category:Primitives involving Root of x squared minus a squared