nyuuzyou/wikis huggingface dataset:
Primitive of Reciprocal of x by Half Integer Power of a x squared plus b x plus c

== Theorem ==
Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n + \frac 1 2} } = \frac 1 {\paren {2 n - 1} c \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c}^{n - \frac 1 2} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }$ 
== Proof ==
Now take:
Let:
Thus:
So:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.298$
* : $\S 17$: Tables of Special Indefinite Integrals: $(13)$ Integrals Involving $\sqrt {a x^2 + b x + c}$: $17.13.19.$
Category:Primitives involving Root of a x squared plus b x plus c
Primitive of Reciprocal of x by Logarithm of x

== Theorem ==
:$\ds \int \frac {\d x} {x \ln x} = \ln \size {\ln x} + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $\ln x$: $14.532$
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitives involving Logarithm Function
Primitive of Reciprocal of x by Power of Root of a x + b

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {\sqrt {a x + b} }^m} = \frac 2 {\paren {m - 2} b \paren {\sqrt {a x + b} }^{m - 2} } + \frac 1 b \int \frac {\d x} {x \paren {\sqrt {a x + b} }^{m - 2} }$
== Proof ==
From Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $a x + b$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$
Putting $n := -\dfrac m 2$ and $m := -1$:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt{a x + b}$: $14.104$
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.21.$
Category:Primitives involving Root of a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x by Power of a squared minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {a^2 - x^2}^n} = \frac 1 {2 \paren {n - 1} a^2 \paren {a^2 - x^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {a^2 - x^2}^{n - 1} }$
for $x^2 < a^2$.
== Proof ==
Let:
From Primitive of $x^m \paren {a x + b}^n$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Let:
Then:
== Also see ==
* Primitive of $\dfrac 1 {x \paren {x^2 - a^2}^n}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.179$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of x by Power of x minus Power of a

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^n - a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n - a^n} {x^n} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.330$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Reciprocal of x by Power of x plus Power of a

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.325$
* : $\S 17$: Tables of Special Indefinite Integrals: $(16)$ Integrals Involving $x^n \pm a^n$: $17.16.1.$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Reciprocal of x by Power of x squared minus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^2 - a^2}^n} = \frac {-1} {2 \paren {n - 1} a^2 \paren {x^2 - a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x \paren {x^2 - a^2}^{n - 1} }$
for $x^2 > a^2$.
== Proof ==
Let:
From Primitive of $x^m \paren {a x + b}^n$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Let:
Then:
== Also see ==
* Primitive of $\dfrac 1 {x \paren {a^2 - x^2}^n}$
== Sources ==
* : $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.160$
* : $\S 17$: Tables of Special Indefinite Integrals: $(7)$ Integrals Involving $x^2 - a^2$, $x^2 > a^2$: $17.7.17.$
Category:Primitives involving x squared minus a squared
Primitive of Reciprocal of x by Power of x squared plus a squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {x^2 + a^2}^n} = \frac 1 {2 \paren {n - 1} a^2 \paren {x^2 + a^2}^{n - 1} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {x^2 + a^2}^{n - 1} }$
== Proof ==
Let:
From Primitive of $x^m \paren {a x + b}^n$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Let:
Then:
== Sources ==
* : $\S 14$: Integrals involving $x^2 + a^2$: $14.141$
* : $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.17.$
Category:Primitives involving x squared plus a squared
Primitive of Reciprocal of x by Root of 2 a x minus x squared

== Theorem ==
:$\ds \int \dfrac 1 {x \sqrt {2 a x - x^2} } \rd x = -\dfrac 1 a \sqrt {\dfrac {2 a - x} x} + C$
== Proof ==
Recall Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$:
Then:
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $55$.
Category:Primitives of Roots of Quadratic Functions
Primitive of Reciprocal of x by Root of Power of x minus Power of a

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^n - a^n} } = \frac 2 {n \sqrt {a^n} } \arccos \sqrt {\frac {a^n} {x^n} }$
== Proof ==
Now:
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.334$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Reciprocal of x by Root of Power of x plus Power of a

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^n + a^n} } = \frac 1 {n \sqrt {a^n} } \ln \size {\frac {\sqrt {x^n + a^n} - \sqrt {a^n} } {\sqrt {x^n + a^n} + \sqrt {a^n} } } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $x^n \pm a^n$: $14.329$
Category:Primitives involving Power of x plus or minus Power of a
Primitive of Reciprocal of x by Root of a squared minus x squared

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

== Theorem ==
For $a > 0$ and $0 < \size x < a$:
:$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {\size x} a} + C$
== Proof ==
We note that $\sech^{-1} \dfrac x a$ is defined for $x \in \hointl 0 a$.
Hence we treat the two cases $x > 0$ and $x < 0$ separately.
First let $x > 0$.
Note that $\dfrac 1 {x \sqrt {a^2 - x^2} }$ is not defined at $\pm a$, so we are concerned only about the interval $\openint 0 a$.
Then:
Now let $x < 0$.
Let $z = -x$.
Then:
So when $x > 0$:
:$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac x a} + C$
and when $x < 0$:
:$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {-x} a} + C$
It follows that for $\size x \in \openint 0 a$:
:$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac {\size x} a} + C$
by definition of absolute value.
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$
Category:Expressions whose Primitives are Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Secant
Category:Primitive of Reciprocal of x by Root of a squared minus x squared
Primitive of Reciprocal of x by Root of a squared minus x squared/Logarithm Form

== Theorem ==
For $a > 0$ and $0 < \size x < a$:
:$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== Proof ==
== Also presented as ==
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.46$
* : $\S 14$: General Rules of Integration: $14.47$
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.241$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.47.$
Category:Primitive of Reciprocal of x by Root of a squared minus x squared
Primitive of Reciprocal of x by Root of a squared minus x squared/Logarithm Form/Also presented as

== Primitive of $\frac 1 {x \sqrt {a^2 - x^2} }$: Logarithm Form: Also presented as ==
This result is also seen presented in the form:
:$\ds \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \ln \size {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $34$.
Category:Primitive of Reciprocal of x by Root of a squared minus x squared
Primitive of Reciprocal of x by Root of a squared minus x squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {\sqrt {a^2 - x^2} }^3} = \frac 1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \map \ln {\frac {a + \sqrt {a^2 - x^2} } {\size x} } + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac 1 {x \paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac 1 {x \paren {\sqrt {x^2 - a^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a^2 - x^2}$: $14.255$
Category:Primitives involving Root of a squared minus x squared
Primitive of Reciprocal of x by Root of a x + b

== Theorem ==
For $a > 0$ and for $x \ne 0$:
:$\ds \int \frac {\d x} {x \sqrt {a x + b} } = \begin {cases}
\dfrac 1 {\sqrt b} \ln \size {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C & : b > 0 \\ \\
\dfrac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C & : b < 0 \end {cases}$
where $a x + b > 0$.
== Proof ==
Let:
Thus:
Then:
Let $b > 0$.
Let $d = \sqrt b$.
Then:
Let $b < 0$.
Let $d = \sqrt {-b}$.
Then:
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.87$
* : Front endpapers: A Brief Table of Integrals: $13$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(2)$ Integrals Involving $\sqrt {a x + b}$: $17.2.4.$
Category:Primitives involving Root of a x + b
Primitive of Reciprocal of x by Root of a x squared plus b x plus c

== Theorem ==
Let $a, b, c \in \R_{\ne 0}$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$ and $x \ne 0$:
$\quad \ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } = \begin {cases}
\dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x} + C & : c > 0, b^2 - 4 a c > 0 \\
\dfrac {-1} {\sqrt c} \map {\sinh^{-1} } {\dfrac {b x + 2 c} {\size x \sqrt {4 a c - b^2} } } + C & : c > 0, b^2 - 4 a c < 0 \\
\dfrac {-1} {\sqrt c} \dfrac {\size x} x \ln \size {\dfrac {2 c} x + b} + C & : c > 0, b^2 - 4 a c = 0 \\
\dfrac 1 {\sqrt {-c} } \map \arcsin {\dfrac {b x + 2 c} {\size x \sqrt {\size {b^2 - 4 a c} } } } & : c < 0, b^2 - 4 a c \ne 0 \\
\end {cases}$
== Proof 1 ==
== Proof 2 (incomplete) ==
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {a x^2 + b x + c}$: $14.283$
Category:Primitives involving Root of a x squared plus b x plus c
Category:Primitive of Reciprocal of x by Root of a x squared plus b x plus c
Primitive of Reciprocal of x by Root of a x squared plus b x plus c/Lemma

== Theorem ==
Let $a, b, c \in \R_{\ne 0}$.
Then for $x \in \R$ such that $a x^2 + b x + c > 0$ and $x \ne 0$:
:$\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } = -\int \frac {\d u} {\pm \sqrt {a + b u + c u^2} }$
where $u := \dfrac 1 x$, according to whether $u > 0$ or $u < 0$.
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.38$
Category:Primitive of Reciprocal of x by Root of a x squared plus b x plus c
Primitive of Reciprocal of x by Root of a x squared plus b x plus c/Proof 1

== Theorem ==
== Proof ==
=== Lemma ===
Let $x > 0$, and so $u > 0$.
Then we have:
:$\ds \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } = -\int \frac {\d u} {\sqrt {a + b u + c u^2} }$
We consider the two cases where $c > 0$ and $c < 0$.
First we take $c > 0$.
Thus:
Then we take $c < 0$.
Thus:
Now we consider what happens when $x < 0$, and so $u < 0$.
We have:
which leads us to:
Finally we take $c < 0$.
Thus:
== Sources ==
* : $\text {II}$. Calculus: Integration: Useful substitutions
Category:Primitive of Reciprocal of x by Root of a x squared plus b x plus c
Primitive of Reciprocal of x by Root of x squared minus a squared

== Theorem ==
=== Arccosine Form ===
=== Arcsecant Form ===
=== Arcsine Form ===
Category:Expressions whose Primitives are Inverse Trigonometric Functions
Category:Primitives involving Root of x squared minus a squared
Category:Primitives involving Reciprocals
Category:Primitive of Reciprocal of x by Root of x squared minus a squared
Primitive of Reciprocal of x by Root of x squared minus a squared/Arccosine Form

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arccos \size {\frac a x} + C$
for $0 < a < \size x$.
== Proof ==
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
:$x > a$
or:
:$x < -a$
where it is assumed that $a > 0$.
Hence:
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.43$
* : Front endpapers: A Brief Table of Integrals: $45$.
Category:Primitive of Reciprocal of x by Root of x squared minus a squared
Category:Primitives involving Inverse Cosine Function
Primitive of Reciprocal of x by Root of x squared minus a squared/Arcsecant Form

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arcsec \size {\frac x a} + C$
for $0 < a < \size x$.
== Proof ==
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
:$x > a$
or:
:$x < -a$
where it is assumed that $a > 0$.
Consider the arcsecant substitution:
:$u = \arcsec {\dfrac x a}$
which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.
That is:
:$\size x \ge a$
and it is seen that $u = \arcsec {\dfrac x a}$ is defined over the whole domain of the integrand.
Hence:
Let $x > a$.
Now suppose $x < -a$.
Let $z = -x$.
Then:
:$\d x = -\d z$
and we then have:
The result follows.
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
== Sources ==
* : $\S 14$: General Rules of Integration: $14.45$
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.213$
* : Front endpapers: A Brief Table of Integrals: $45$.
* : Chapter $25$: Fundamental Integration Formulas: $20$.
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.45.$
Category:Primitive of Reciprocal of x by Root of x squared minus a squared
Category:Arcsecant Function
Primitive of Reciprocal of x by Root of x squared minus a squared/Arcsine Form

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = -\frac 1 a \arcsin \size {\frac a x} + C$
for $0 < a < \size x$.
== Proof ==
We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
:$x > a$
or:
:$x < -a$
where it is assumed that $a > 0$.
Hence:
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
== Sources ==
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Arcsine Function
Category:Primitive of Reciprocal of x by Root of x squared minus a squared
Primitive of Reciprocal of x by Root of x squared minus a squared/Arcsine Form/Mistake

== Source Work ==
:Table 2 Integrals
== Mistake ==
:$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = -\frac 1 a \sin^{-1} \frac a x + C$
== Correction ==
Should be:
:$\ds \int \frac {\d x} {x \sqrt {x^2 - a^2} } = -\frac 1 a \arcsin \size {\frac a x} + C$
It also needs to specify that the result is valid only where $x^2 > a^2$.
== Sources ==
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Mistakes/The Penguin Dictionary of Mathematics
Primitive of Reciprocal of x by Root of x squared minus a squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3} = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} + C$
for $\size x > a$.
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac 1 {x \paren {\sqrt {x^2 + a^2} }^3}$
* Primitive of $\dfrac 1 {x \paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 - a^2}$: $14.227$
Category:Primitives involving Root of x squared minus a squared
Primitive of Reciprocal of x by Root of x squared plus a squared

== Theorem ==
=== Inverse Hyperbolic Cosecant Form ===
=== Inverse Hyperbolic Sine Form ===
=== Logarithm Form ===
=== Reciprocal Logarithm Form ===
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
Category:Primitives involving Root of x squared plus a squared
Category:Primitives involving Reciprocals
Category:Primitive of Reciprocal of x by Root of x squared plus a squared
Primitive of Reciprocal of x by Root of x squared plus a squared/Inverse Hyperbolic Cosecant Form

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \csch^{-1} {\frac x a} + C$
== Proof ==
Let:
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
Category:Primitive of Reciprocal of x by Root of x squared plus a squared
Category:Expressions whose Primitives are Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Cosecant
Primitive of Reciprocal of x by Root of x squared plus a squared/Inverse Hyperbolic Sine Form

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \sinh^{-1} {\frac a x} + C$
== Proof ==
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
== Sources ==
* : Appendix: Table $2$: Integrals
* : Appendix: Table $2$: Integrals
Category:Primitive of Reciprocal of x by Root of x squared plus a squared
Category:Expressions whose Primitives are Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Sine
Primitive of Reciprocal of x by Root of x squared plus a squared/Logarithm Form

== Theorem ==
For $x \in \R_{\ne 0}$:
:$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \map \ln {\frac a x + \frac {\sqrt {a^2 + x^2} } {\size x} } + C$
== Proof ==
== Also presented as ==
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
== Sources ==
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions: $3.3.42$
* : $\S 14$: General Rules of Integration: $14.46$
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.186$
* : $\S 16$: Indefinite Integrals: General Rules of Integration: $16.46.$
Category:Primitive of Reciprocal of x by Root of x squared plus a squared
Primitive of Reciprocal of x by Root of x squared plus a squared/Logarithm Form/Also presented as

== Primitive of $\frac 1 {x \sqrt {x^2 + a^2} }$: Logarithm Form: Also presented as ==
This result is also seen presented in the form:
:$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \ln \size {\frac a x + \frac {\sqrt {a^2 + x^2} } x} + C$
== Sources ==
* : Front endpapers: A Brief Table of Integrals: $26$.
Category:Primitive of Reciprocal of x by Root of x squared plus a squared
Primitive of Reciprocal of x by Root of x squared plus a squared/Reciprocal Logarithm Form

== Theorem ==
:$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} } = \frac 1 a \map \ln {\frac x {a + \sqrt {x^2 + a^2} } } + C$
== Proof ==
== Also see ==
* Primitive of Reciprocal of $x \sqrt {x^2 - a^2}$
* Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$
== Sources ==
* : Chapter $5$. Integrals: Integrals of Special Functions: $29$
Category:Primitive of Reciprocal of x by Root of x squared plus a squared
Primitive of Reciprocal of x by Root of x squared plus a squared cubed

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {\sqrt {x^2 + a^2} }^3} = \frac 1 {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
== Proof ==
Let:
== Also see ==
* Primitive of $\dfrac 1 {x \paren {\sqrt {x^2 - a^2} }^3}$
* Primitive of $\dfrac 1 {x \paren {\sqrt {a^2 - x^2} }^3}$
== Sources ==
* : $\S 14$: Integrals involving $\sqrt {x^2 + a^2}$: $14.200$
Category:Primitives involving Root of x squared plus a squared
Primitive of Reciprocal of x by a squared minus x squared

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {a^2 - x^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x \paren {x^2 - a^2} }$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.167$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of x by a squared minus x squared/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\d x} {x \paren {a^2 - x^2}^2} = \frac 1 {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.
== Proof ==
== Also see ==
* Primitive of $\dfrac 1 {x \paren {x^2 - a^2}^2}$
== Sources ==
* : $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.174$
Category:Primitives involving a squared minus x squared
Primitive of Reciprocal of x by a squared minus x squared squared/Partial Fraction Expansion

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

== Theorem ==
:$\ds \int \frac {\rd x} {x \paren {a x + b} } = \frac 1 b \ln \size {\frac x {a x + b} } + C$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.63$
* : Front endpapers: A Brief Table of Integrals: $10$.
* : $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.4.$
Category:Primitive of Reciprocal of x by a x + b‎
Category:Primitives involving a x + b
Category:Primitives involving Reciprocals
Primitive of Reciprocal of x by a x + b/Partial Fraction Expansion

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

== Theorem ==
:$\ds\int \frac {\d x} {x \paren {a x + b}^3} = \frac {a^2 x^2} {2 b^3 \paren {a x + b}^2} - \frac {2 a x} {b^3 \paren {a x + b} } + \frac 1 {b^3} \ln \size {\frac x {a x + b} } + C$
== Proof ==
== Sources ==
* : $\S 14$: Integrals involving $a x + b$: $14.77$
Category:Primitives involving a x + b‎