avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Power of x by Arcsecant of x

:$\ds \int x^m \arcsec x \rd x = \begin {cases}
\dfrac {x^{m + 1} } {m + 1} \arcsec x - \dfrac 1 {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arcsec x < \dfrac \pi 2 \\
\dfrac {x^{m + 1} } {m + 1} \arcsec x + \dfrac 1 {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : \dfrac \pi 2 < \arcsec x < \pi \\
\end {cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = <nowiki> \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \\
\dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \\
\end {cases} </nowiki>
      | c = Derivative of $\arcsec \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^{m + 1} } {m + 1}
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arcsec \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arcsec \frac x a \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arcsec \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arcsec \frac x a - \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Similarly, let $\arcsec \dfrac x a$ be in the interval $\openint {\dfrac \pi 2} \pi$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arcsec \frac x a \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arcsec \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arcsec \frac x a + \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arccosecant of x

:$\ds \int x^m \arccsc x \rd x = \begin {cases}
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x + \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x - \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\
\end {cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\rd v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccsc x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\
\dfrac 1 {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\
\end{cases}
      | c = Derivative of $\arccsc x$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^{m + 1} } {m + 1}
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arccsc x$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arccsc x \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arccsc x - \int \frac {x^{m + 1} } {m + 1} \paren {\frac {-1} {x \sqrt {x^2 - 1} } } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arccsc x + \frac 1 {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - 1} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2}  0$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arccsc x \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arccsc x - \int \frac {x^{m + 1} } {m + 1} \paren {\frac 1 {x \sqrt {x^2 - 1} } } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arccsc x - \frac 1 {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - 1} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arccosecant of x

:$\ds \int x^m \arccsc x \rd x = \begin {cases}
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x + \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x - \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\
\end {cases}$

Proof:
Recall:
{{begin-eqn}}
{{eqn | n = 1
      | l = \int x^m \arccsc x \rd x
      | r = <nowiki>\begin {cases}
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x + \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\ \\
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x - \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\
\end {cases} </nowiki>
      | c = Primitive of $x^m \arccsc x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arccsc \frac x a \rd x
      | r = \int a^m \paren {\dfrac x a}^m \arccsc \frac x a \rd x
      | c = manipulating into appropriate form
}}
{{eqn | r = a^m \int \paren {\dfrac x a}^m \arccsc \frac x a \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = <nowiki> a^m \paren {\dfrac 1 {1 / a} \paren {\begin {cases}
\dfrac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arccsc \dfrac x a + \dfrac 1 {m + 1} \ds \int \paren {\dfrac x a}^m \frac {\d x} {\sqrt {\paren {\dfrac x a}^2 - 1} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \\
\dfrac 1 {m + 1} \paren {\dfrac x a}^{m + 1} \arccsc \dfrac x a - \dfrac 1 {m + 1} \ds \int \paren {\dfrac x a}^m \frac {\d x} {\sqrt {\paren {\dfrac x a}^2 - 1} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0
\end {cases} } } </nowiki>
      | c = Primitive of Function of Constant Multiple, from $(1)$
}}
{{eqn | r = <nowiki> \begin {cases}
\dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a + \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\ \\
\dfrac {x^{m + 1} } {m + 1} \arccsc \dfrac x a - \dfrac a {m + 1} \ds \int \dfrac {x^m \rd x} {\sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0
\end {cases} </nowiki>
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arccosecant of x

:$\ds \int x^m \arccsc x \rd x = \begin {cases}
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x + \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \\
\ds \dfrac {x^{m + 1} } {m + 1} \arccsc x - \dfrac 1 {m + 1} \int \dfrac {x^m \rd x} {\sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \\
\end {cases}$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\rd v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arccsc \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = <nowiki> \begin {cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \arccsc \dfrac x a < \dfrac \pi 2 \\
\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\
\end{cases} </nowiki>
      | c = Derivative of $\arccsc \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^{m + 1} } {m + 1}
      | c = Primitive of Power
}}
{{end-eqn}}
First let $\arccsc \dfrac x a$ be in the interval $\openint 0 {\dfrac \pi 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arccsc \frac x a \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arccsc \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac {-a} {x \sqrt {x^2 - a^2} } } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arccsc \frac x a + \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Similarly, let $\arccsc \dfrac x a$ be in the interval $\openint {-\dfrac \pi 2}  0$.
Then:
{{begin-eqn}}
{{eqn | l = \int x^m \arccsc \frac x a \rd x
      | r = \frac {x^{m + 1} } {m + 1} \arccsc \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac a {x \sqrt {x^2 - a^2} } } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{m + 1} } {m + 1} \arccsc \frac x a - \frac a {m + 1} \int \frac {x^m \rd x} {\sqrt {x^2 - a^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arcsine of a x

:$\ds \int x^n \arcsin a x \rd x = \frac {x^{n + 1} } {n + 1} \arcsin a x - \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} }$
for $n \ne -1$.

Proof:
Recall:
{{begin-eqn}}
{{eqn | n = 1
      | l = \int x^n \arcsin x \rd x
      | r = \frac {x^{n + 1} } {n + 1} \arcsin x - \frac 1 {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - x^2} }
      | c = Primitive of $x^n \arcsin x$
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = u 
      | r = a x
}}
{{eqn | ll = \leadsto
      | l = \rd u
      | r = a \rd x
      | c = differentiating {{WRT|Differentiation}} $x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^n \arcsin a x \rd x
      | r = \int \paren {\frac u a}^n \arcsin u \paren {\frac 1 a \rd u }
      | c = Integration by Substitution
}}
{{eqn | r = \dfrac 1 {a^{n + 1} } \int u^n \arcsin u \rd u
      | c = 
}}
{{eqn | r = \dfrac 1 {a^{n + 1} } \paren {\frac {u^{n + 1} } {n + 1} \arcsin u - \frac 1 {n + 1} \int \frac {u^{n + 1} \rd u} {\sqrt {1 - u^2} } }
      | c = from $\paren 1$ above
}}
{{eqn | r = \frac {\paren {\frac u a}^{n + 1} } {n + 1} \arcsin u - \frac 1 {n + 1} \int \frac {\paren {\frac u a}^{n + 1} \paren {a \rd x } } {\sqrt {1 - u^2} }
      | c = $\rd u = a \rd x$
}}
{{eqn | r = \frac {x^{n + 1} } {n + 1} \arcsin a x - \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} } 
      | c = $\dfrac u a = x$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arccosine of a x

:$\ds \int x^n \arccos a x \rd x = \frac {x^{n + 1} } {n + 1} \arccos a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} }$
for $n \ne -1$.

Proof:
Recall:
{{begin-eqn}}
{{eqn | n = 1
      | l = \int x^n \arccos x \rd x
      | r = \frac {x^{n + 1} } {n + 1} \arccos x + \frac 1 {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - x^2} }
      | c = Primitive of $x^n \arccos x$
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = u 
      | r = a x
}}
{{eqn | ll = \leadsto
      | l = \rd u
      | r = a \rd x
      | c = differentiating {{WRT|Differentiation}} $x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^n \arccos a x \rd x
      | r = \int \paren {\frac u a}^n \arccos u \paren {\frac 1 a \rd u }
      | c = Integration by Substitution
}}
{{eqn | r = \dfrac 1 {a^{n + 1} } \int u^n \arccos u \rd u
      | c = 
}}
{{eqn | r = \dfrac 1 {a^{n + 1} } \paren {\frac {u^{n + 1} } {n + 1} \arccos u + \frac 1 {n + 1} \int \frac {u^{n + 1} \rd u} {\sqrt {1 - u^2} } }
      | c = from $\paren 1$ above
}}
{{eqn | r = \frac {\paren {\frac u a}^{n + 1} } {n + 1} \arccos u + \frac 1 {n + 1} \int \frac {\paren {\frac u a}^{n + 1} \paren {a \rd x} } {\sqrt {1 - u^2} }
      | c = $\rd u = a \rd x$
}}
{{eqn | r = \frac {x^{n + 1} } {n + 1} \arccos a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {\sqrt {1 - a^2 x^2} } 
      | c = $\dfrac u a = x$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Arctangent of a x

:$\ds \int x^n \arctan a x \rd x = \frac {x^{n + 1} } {n + 1} \arctan a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {a^2 x^2 + 1}$
for $n \ne -1$.

Proof:
Recall:
{{begin-eqn}}
{{eqn | n = 1
      | l = \int x^n \arctan x \rd x
      | r = \frac {x^{n + 1} } {n + 1} \arctan x - \frac 1 {n + 1} \int \frac {x^{n + 1} \rd x} {x^2 + 1}
      | c = Primitive of $x^n \arctan x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^n \arctan a x \rd x
      | r = \int \dfrac 1 {a^n} \paren {a x}^n \arctan a x \rd x
      | c = manipulating into appropriate form
}}
{{eqn | r = \dfrac 1 {a^n} \int \paren {a x}^n \arctan a x \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \dfrac 1 {a^n} \paren {\dfrac 1 a \paren {\frac {\paren {a x}^{n + 1} } {n + 1}  \arctan a x + \frac 1 {n + 1} \int \frac {\paren {a x}^{n + 1} \rd x} {\paren {a x}^2 + 1} } }
      | c = Primitive of Function of Constant Multiple, from $(1)$
}}
{{eqn | r = \frac {x^{n + 1} } {n + 1} \arctan a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {a^2 x^2 + 1} 
      | c = simplifying {{Mistake|can't resolve the $\dfrac a {n + 1}$ factor, it seems to want to be $\dfrac 1 {n + 1}$}}
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of General Exponential of a x

:$\ds \int b^{a x} \rd x = \frac {b^{a x} } {a \ln b} + C$
where:
:$a \ne 0$
:$b > 0, b \ne 1$

Proof:
{{begin-eqn}}
{{eqn | l = \int b^x \rd x
      | r = \frac {b^x} {\ln b} + C
      | c = Primitive of $b^x$
}}
{{eqn | ll= \leadsto
      | l = \int b^{a x} \rd x
      | r = \frac 1 a \paren {\frac {b^x} {\ln b} } + C
      | c = Primitive of Function of Constant Multiple
}}
{{eqn | r = \frac {b^{a x} } {a \ln b} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Reduction Formula for Power of x by Exponential of a x

:$\ds \int x^n e^{a x} \rd x = \frac {x^n e^{a x} } a - \dfrac n a \int x^{n - 1} e^{a x} \rd x$
for $n \in \Z_{>0}$, $a \in \R_{\ne 0}$.

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = x^n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = n x^{n - 1}
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = e^{a x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \dfrac {e^{a x} } a
      | c = Primitive of $e^{a x}$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^n e^{a x} \rd x
      | r = \int u \rd v
      | c = 
}}
{{eqn | r = \paren {x^n} \paren {\dfrac {e^{a x} } a} - \int \paren {\dfrac {e^{a x} } a} \paren {n x^{n - 1} } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^n e^{a x} } a - \dfrac n a \int x^{n - 1} e^{a x} \rd x
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Reduction Formula for Power of x by General Exponential of a x

:$\ds \int x^n b^{a x} \rd x = \frac {x^n b^{a x} } {a \ln b} - \dfrac n {a \ln b} \int x^{n - 1} b^{a x} \rd x$
for $n \in \Z_{>0}$, $a \in \R_{\ne 0}$, $b > 0$, $b \ne 1$

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = x^n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = n x^{n - 1}
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = b^{a x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \dfrac {e^{a x} } {a \ln b}
      | c = Primitive of $b^{a x}$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^n b^{a x} \rd x
      | r = \int u \rd v
      | c = 
}}
{{eqn | r = \paren {x^n} \paren {\dfrac {b^{a x} } {a \ln b} } - \int \paren {\dfrac {b^{a x} } {a \ln b} } \paren {n x^{n - 1} } \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^n b^{a x} } {a \ln b} - \dfrac n {a \ln b} \int x^{n - 1} b^{a x} \rd x
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Logarithm of a x

:$\ds \int \ln a x \rd x = x \ln a x - x + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \ln x \rd x
      | r = x \ln x - x + C
      | c = Primitive of $\ln x$
}}
{{eqn | ll= \leadsto
      | l = \int \ln a x \rd x
      | r = \frac 1 a \paren {a x \ln a x - a x} + C
      | c = Primitive of Function of Constant Multiple
}}
{{eqn | r = x \ln a x - x + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Derivative of Natural Logarithm of a x

Let $\ln x$ be the natural logarithm function.
Then:
:$\map {\dfrac \d {\d x} } {\ln a x} = \dfrac 1 x$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\ln a x}
      | r = a \map {\dfrac \d {\d \paren {a x} } } {\ln a x}
      | c = Derivative of Function of Constant Multiple
}}
{{eqn | r = a \dfrac 1 {a x}
      | c = Derivative of Natural Logarithm
}}
{{eqn | r = \dfrac 1 x
      | c = 
}}
{{end-eqn}}
Category:Derivatives involving Logarithm Functions
47f6rnr02e8ynfwgu77dkf4rtdzvxnc
Theorem: Primitive of Power of x by Logarithm of a x

:$\ds \int x^n \ln a x \rd x = \dfrac {x^{n + 1} } {n + 1} \ln a x - \dfrac {x^{n + 1} } {\paren {n + 2}^2} + C$
for $n \ne -1$.

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \ln a x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 x
      | c = Derivative of $\ln a x$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = x^n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {x^{n + 1} } {n + 1}
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x^n \ln a x \rd x
      | r = \frac {x^{n + 1} } {n + 1} \ln a x - \int \frac {x^{n + 1} } {n + 1} \paren {\frac 1 x} \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac {x^{n + 1} } {n + 1} \ln a x - \frac 1 {n + 1} \int x^n \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac {x^{n + 1} } {n + 1} \ln a x - \frac 1 {n + 1} \paren {\frac {x^{n + 1} } {n + 1} } + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {x^{n + 1} } {n + 1} \paren {\ln a x - \frac 1 {n + 1} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Logarithm of a x

:$\ds \int \frac {\d x} {x \ln a x} = \ln \size {\ln a x} + C$

Proof:
{{begin-eqn}}
{{eqn | l = z
      | r = \ln a x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = \frac 1 x
      | c = Derivative of $\ln a x$
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x \ln a x}
      | r = \ln \size {\ln a x} + C
      | c = Primitive of Function under its Derivative
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Exponential of a x by Hyperbolic Sine of b x/Exponential Form

:$\ds \int e^{a x} \sinh b x \rd x = \frac {e^{a x} } 2 \paren {\frac {e^{b x} } {a + b} - \frac {e^{-b x} } {a - b} } + C$
for $a^2 \ne b^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int e^{a x} \sinh b x \rd x
      | r = \int e^{a x} \paren {\frac {e^{b x} - e^{-b x} } 2} \rd x
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac 1 2 \int e^{a x} \paren {e^{b x} - e^{-b x} } \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \int \paren {e^{a x} e^{b x} - e^{a x} e^{-b x} } \rd x
      | c = 
}}
{{eqn | r = \frac 1 2 \int \paren {e^{a x + b x} - e^{a x - b x} } \rd x
      | c = Product of Powers
}}
{{eqn | r = \frac 1 2 \int e^{a x + b x} \rd x - \frac 1 2 \int e^{a x - b x} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 2 \int e^{\paren {a + b} x} \rd x - \frac 1 2 \int e^{\paren {a - b} x} \rd x
      | c = 
}}
{{eqn | r = \frac 1 2 \frac {e^{\paren {a + b} x} } {a + b} - \frac 1 2 \frac {e^{\paren {a - b} x} } {a - b} + C
      | c = Primitive of $e^{a x}$
}}
{{eqn | r = \frac 1 2 \frac {e^{a x + b x} } {a + b} - \frac 1 2 \frac {e^{a x - b x} } {a - b} + C
      | c = 
}}
{{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} } {a + b} - \frac 1 2 \frac {e^{a x} e^{-b x} } {a - b} + C
      | c = Product of Powers
}}
{{eqn | r = \frac {e^{a x} } 2 \paren {\frac {e^{b x} } {a + b} - \frac {e^{-b x} } {a - b} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Exponential of a x by Hyperbolic Sine of b x/Hyperbolic Form

:$\ds \int e^{a x} \sinh b x \rd x = \frac {e^{a x} \paren {a \sinh b x - b \cosh b x} } {a^2 - b^2} + C$
for $a^2 \ne b^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int e^{a x} \sinh b x \rd x
      | r = \frac {e^{a x} } 2 \paren {\frac {e^{b x} } {a + b} - \frac {e^{-b x} } {a - b} } + C
      | c = Primitive of $e^{a x} \sinh b x$: Exponential Form
}}
{{eqn | r = \frac {e^{a x} } 2 \paren {\frac {e^{b x} \paren {a - b} } {\paren {a + b} \paren {a - b} } - \frac {e^{-b x} \paren {a + b} } {\paren {a - b} \paren {a + b} } } + C
      | c = 
}}
{{eqn | r = \frac {e^{a x} } 2 \paren {\frac {e^{b x} \paren {a - b} - e^{-b x} \paren {a + b} } {\paren {a + b} \paren {a - b} } } + C
      | c = 
}}
{{eqn | r = \frac {e^{a x} } 2 \paren {\frac {e^{b x} \paren {a - b} - e^{-b x} \paren {a + b} } {a^2 - b^2} } + C
      | c = Difference of Two Squares
}}
{{eqn | r = \frac {e^{a x} } {a^2 - b^2} \paren {\frac {a e^{b x} - b e^{b x} - a e^{-b x} - b e^{-b x} } 2} + C
      | c = 
}}
{{eqn | r = \frac {e^{a x} } {a^2 - b^2} \paren {a \frac {e^{b x} - e^{-b x} } 2 - b \frac {e^{b x} + e^{-b x} } 2} + C
      | c = 
}}
{{eqn | r = \frac {e^{a x} } {a^2 - b^2} \paren {a \frac {e^b x - e^{-b} x} 2 - b \cosh b x} + C
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac {e^{a x} } {a^2 - b^2} \paren {a \sinh b x - b \cosh b x} + C
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac {e^{a x} \paren {a \sinh b x - b \cosh b x} } {a^2 - b^2} + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Exponential of a x by Hyperbolic Cosine of b x/Exponential Form

:$\ds \int e^{a x} \cosh b x \rd x = \frac {e^{a x} } 2 \paren {\frac {e^{b x} } {a + b} + \frac {e^{-b x} } {a - b} } + C$
for $a^2 \ne b^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int e^{a x} \cosh b x \rd x
      | r = \int e^{a x} \paren {\frac {e^{b x} + e^{-b x} } 2} \rd x
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac 1 2 \int e^{a x} \paren {e^{b x} + e^{-b x} } \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \int \paren {e^{a x} e^{b x} + e^{a x} e^{-b x} } \rd x
      | c = 
}}
{{eqn | r = \frac 1 2 \int \paren {e^{a x + b x} + e^{a x - b x} } \rd x
      | c = Product of Powers
}}
{{eqn | r = \frac 1 2 \int e^{a x + b x} \rd x + \frac 1 2 \int e^{a x - b x} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 2 \int e^{\paren {a + b} x} \rd x + \frac 1 2 \int e^{\paren {a - b} x} \rd x
      | c = 
}}
{{eqn | r = \frac 1 2 \frac {e^{\paren {a + b} x} } {a + b} + \frac 1 2 \frac {e^{\paren {a - b} x} } {a - b} + C
      | c = Primitive of $e^{a x}$
}}
{{eqn | r = \frac 1 2 \frac {e^{a x + b x} } {a + b} + \frac 1 2 \frac {e^{a x - b x} } {a - b} + C
      | c = 
}}
{{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} } {a + b} + \frac 1 2 \frac {e^{a x} e^{-b x} } {a - b} + C
      | c = Product of Powers
}}
{{eqn | r = \frac {e^{a x} } 2 \paren {\frac {e^{b x} } {a + b} + \frac {e^{-b x} } {a - b} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}