avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Root of a x + b

:$\ds \int \sqrt {a x + b} \rd x = \frac {2 \sqrt {\paren {a x + b}^3} } {3 a}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = v
      | r = \sqrt x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = \frac 1 {2 \sqrt x}
      | c = Power Rule for Derivatives
}}
{{eqn | l = u
      | r = \frac {2 \sqrt {a x + b} } {\sqrt x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {\frac {\sqrt x \cdot 2 a} {2 \sqrt{a x + b} } - \frac {2 \sqrt {a x + b} } {2 \sqrt x} } x
      | c = Quotient Rule for Derivatives etc.
}}
{{eqn | r = \frac {-b} {x^{3/2} \sqrt {a x + b} }
      | c = simplifying
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } x \rd x
      | r = \int \frac {2 \sqrt {a x + b} }{\sqrt x} \frac 1 {2 \sqrt x} \rd x
      | c = 
}}
{{eqn | r = \frac {2 \sqrt {a x + b} } {\sqrt x} \sqrt x - \int {\sqrt x} \frac {-b} {x^{3/2} \sqrt {a x + b} } \rd x
      | c = 
}}
{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form

:$\ds \int \frac {\d x} {a^2 - x^2} = \frac 1 a \tanh^{-1} \frac x a + C$
where $\size x < a$.

Proof:
Let $\size x < a$.
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \tanh^{-1} {\frac x a}
      | c = {{Defof|Real Inverse Hyperbolic Tangent}}, which is defined where $\size {\dfrac x a} < 1$
}}
{{eqn | ll=\leadsto
      | l = x
      | r = a \tanh u
      | c = 
}}
{{eqn | ll=\leadsto
      | l = \frac {\d x} {\d u}
      | r = a \sech^2 u
      | c = Derivative of Hyperbolic Tangent
}}
{{eqn | ll=\leadsto
      | l = \int \frac {\rd x} {a^2 - x^2}
      | r = \int \frac {a \sech^2 u} {a^2 - a^2 \tanh^2 u} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac a {a^2} \int \frac {\sech^2 u} {1 - \tanh^2 u} \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 a \int \frac {\sech^2 u} {\sech^2 u} \rd u
      | c = Sum of Squares of Hyperbolic Secant and Tangent
}}
{{eqn | r = \frac 1 a \int \rd u
}}
{{eqn | r = \frac 1 a u + C
      | c = Integral of Constant
}}
{{eqn | r = \frac 1 a \tanh^{-1} \frac x a + C
      | c = Definition of $u$
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal of Root of x squared plus a squared/Inverse Hyperbolic Sine Form

:$\ds \int \frac {\d x} {\sqrt {x^2 + a^2} } = \arsinh {\frac x a} + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arsinh {\frac x a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \sinh u
      | c = {{Defof|Real Area Hyperbolic Sine}}
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = a \cosh u
      | c = Derivative of Hyperbolic Sine
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\sqrt {x^2 + a^2} }
      | r = \int \frac {a \cosh u} {\sqrt {a^2 \sinh^2 u + a^2} } \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac a a \int \frac {\cosh u} {\sqrt {\sinh^2 u + 1} } \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \int \frac {\cosh u} {\cosh u} \rd u
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = \int 1 \rd u
}}
{{eqn | r = u + C
      | c = Integral of Constant
}}
{{eqn | r = \arsinh {\frac x a} + C
      | c = Definition of $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form

:$\ds \int \frac {\d x} {\sqrt {x^2 - a^2} } = \dfrac {\size x} x \arcosh {\size {\frac x a} } + C$
for $x^2 > a^2$.

Proof:
When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.
When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.
Hence the domain needs to be restricted to $\size x > a$, or that is: $\size {\dfrac x a} > 1$.
Let $x > a$.
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \map \arcosh {\frac x a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \cosh u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = a \sinh u
      | c = Derivative of Hyperbolic Cosine Function
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }
      | r = \int \frac {a \sinh u} {\sqrt {a^2 \cosh^2 u - a^2} } \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac a a \int \frac {\sinh u} {\sqrt {\cosh^2 u - 1} } \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \int \frac {\sinh u} {\sinh u} \rd u
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = \int 1 \rd u
}}
{{eqn | r = u + C
      | c = Integral of Constant
}}
{{eqn | r = \map \arcosh {\frac x a} + C
      | c = Definition of $u$
}}
{{eqn | r = \arcosh \size {\frac x a} + C
      | c = as $\dfrac x a > 0$
}}
{{eqn | r = \dfrac {\size x} x \arcosh \size {\frac x a} + C
      | c = as $x > 0$, so $\dfrac {\size x} x = 1$
}}
{{end-eqn}}
Let $x < -a$.
Let $z = -x$.
Hence:
:$\dfrac {\d z} {\d x} = -1$
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }
      | r = \int \frac {-\d z} {\sqrt {\paren {-z}^2 - a^2} }
      | c = Integration by Substitution
}}
{{eqn | r = -\int \frac {\d z} {\sqrt {z^2 - a^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\map \arcosh {\frac z a} + C
      | c = from above
}}
{{eqn | r = -\map \arcosh {\frac {-x} a} + C
      | c = substituting back
}}
{{eqn | r = -\arcosh \size {\frac x a} + C
      | c = as $\dfrac x a < 0$
}}
{{eqn | r = \dfrac {\size x} x \arcosh \size {\frac x a} + C
      | c = as $x < 0$, so $\dfrac {\size x} x = -1$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Root of x squared plus a squared/Inverse Hyperbolic Cosecant Form

:$\ds \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \csch^{-1} {\frac x a} + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \csch^{-1} {\frac x a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \csch u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = -a \csch u \coth u
      | c = Derivative of Hyperbolic Cosecant
}}
{{eqn | ll = \leadsto
      | l = \int \frac {\d x} {x \sqrt {x^2 + a^2} }
      | r = \int \frac {-a \csch u \coth u} {a \csch u \sqrt {a^2 \csch^2 u + a^2} } \rd u
      | c = Integration by Substitution
}}
{{eqn | r = -\frac a {a^2} \int \frac {\csch u \coth u} {\csch u \sqrt {\csch^2 u - 1} } \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 a \int \frac {\csch u \coth u} {\csch u \coth u} \rd u
      | c = Difference of Squares of Hyperbolic Cotangent and Cosecant
}}
{{eqn | r = -\frac 1 a \int 1 \rd u
}}
{{eqn | r = -\frac 1 a u + C
      | c = Integral of Constant
}}
{{eqn | r = \frac 1 a \csch^{-1} {\frac x a} + C
      | c = Definition of $u$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form

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:
{{begin-eqn}}
{{eqn | l = u
      | r = \sech^{-1} {\frac x a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \sech u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = -a \sech u \tanh u
      | c = Derivative of Hyperbolic Secant
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x \sqrt {a^2 - x^2} }
      | r = \int \frac {-a \sech u \tanh u} {a \sech u \sqrt {a^2 - a^2 \sech^2 u} } \rd u
      | c = Integration by Substitution
}}
{{eqn | r = -\frac a {a^2} \int \frac {\sech u \tanh u} {\sech u \sqrt {1 - \sech^2 u} } \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 a \int \frac {\sech u \tanh u} {\sech u \tanh u} \rd u
      | c = Sum of Squares of Hyperbolic Secant and Tangent
}}
{{eqn | r = -\frac 1 a \int 1 \rd u
}}
{{eqn | r = -\frac 1 a u + C
      | c = Integral of Constant
}}
{{eqn | r = -\frac 1 a \sech^{-1} {\frac x a} + C
      | c = Definition of $u$
}}
{{end-eqn}}
{{qed|lemma}}
Now let $x < 0$.
Let $z = -x$.
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = -x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = -1
      | c = Derivative of Constant Multiple
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x \sqrt {a^2 - x^2} }
      | r = \int \frac {-\d z} {\paren {-z} \sqrt {a^2 - z^2} }
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac {\d z} {z \sqrt {a^2 - z^2} }
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 a \sech^{-1} {\frac z a} + C
      | c = from above: $z \in \hointl 0 a$
}}
{{eqn | r = -\frac 1 a \sech^{-1} {\frac {-x} a} + C
      | c = Definition of $z$
}}
{{end-eqn}}
{{qed|lemma}}
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.
{{qed}}
Theorem: Primitive of x by Root of a x + b

:$\ds \int x \sqrt {a x + b} \rd x = \frac {2 \paren {3 a x - 2 b} } {15 a^2} \sqrt {\paren {a x + b}^3}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \sqrt {a x + b}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {u^2 - b} a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \sqrt {a x + b} \rd x
      | r = \frac 2 a \int \frac {u^2 - b} a u^2 \rd u
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 {a^2} \int u^4 \rd u - \frac {2 b} {a^2} \int u^2 \rd u
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 2 {a^2} \frac {u^5} 5 - \frac {2 b} {a^2} \frac {u^3} 3 + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {6 u^5 - 10 b u^3} {15 a^2} + C
      | c = common denominator
}}
{{eqn | r = \frac {\paren {6 u^2 - 10 b} u^3} {15 a^2} + C
      | c = extracting $u^3$ as a factor
}}
{{eqn | r = \frac {6 \paren {a x + b} - 10 b} {15 a^2} \sqrt {\paren {a x + b}^3} + C
      | c = substituting for $u$
}}
{{eqn | r = \frac {2 \paren {3 a x - 2 b} } {15 a^2} \sqrt {\paren {a x + b}^3} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Heaviside Step Function times Function

Let $\map f t: \R \to \R$ or $\R \to \C$ be a function of exponential order $a$ for some constant $a \in \R$.
Let $f$ be piecewise continuous with one-sided limits on any closed interval of the form $\closedint 0 b$ where $b > 0$.
Let $\map {u_c} t$ be the Heaviside step function.
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.
Then:
:$\laptrans {\map {u_c} t \, \map f {t - c} } = e^{-s c} \map F s$
for $\map \Re s > a$.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\map {u_c} t \, \map f {t - c} }
      | r = \int_0^{\to +\infty} \map {u_c} t e^{-s t} \map f {t - c} \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \int_0^{\to c^-} \map {u_c} t e^{-s t} \map f {t - c} \rd t + \int_{\to c^+}^{\to +\infty} \map {u_c} t e^{-s t} \map f {t - c} \rd t
      | c = Sum of Complex Integrals on Adjacent Intervals
}}
{{eqn | r = \int_0^{\to c^-} 0 \times e^{-s t} \map f {t - c} \rd t + \int_{\to c^+}^{\to +\infty} 1 \times e^{-s t} \map f {t - c} \rd t
      | c = {{Defof|Heaviside Step Function}}
}}
{{eqn | r = \int_{\to c^+}^{\to +\infty} e^{-s t} \map f {t - c} \rd t
}}
{{end-eqn}}
Let $u = t - c$.
Then $\dfrac {\rd u} {\rd t} = 1$.
Then $u \to 0^-$ as $t \to c^+$.
Also, $u \to +\infty$ as $t \to +\infty$.
So:
{{begin-eqn}}
{{eqn | l = \int_{\to c^+}^{\to +\infty} e^{-s t} \map f {t - c} \rd t
      | r = \int_{\to c^+}^{\to +\infty} e^{-s \paren {u + c} } \map f u \frac {\rd u} {\rd t} \rd t
}}
{{eqn | r = \int_{\to 0^-}^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u
      | c = Integration by Substitution
}}
{{eqn | r =  \int_0^{\to 0^+} e^{-s \paren {u + c} } \map f u \rd u + \int_{\to 0^-}^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u
      | c = Integral on Zero Interval
}}
{{eqn | r = \int_0^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u
      | c = Sum of Complex Integrals on Adjacent Intervals
}}
{{eqn | r = \int_0^{\to +\infty} e^{-s u - s c} \map f u \rd u
}}
{{eqn | r = e^{-s c} \int_0^{\to +\infty} e^{-s u} \map f u \rd u
      | c = Exponent Combination Laws
}}
{{eqn | r = e^{-s c} \int_0^{\to +\infty} e^{-s t} \map f t \rd t
      | c = renaming the variable of integration
}}
{{eqn | r = e^{-s c} \laptrans {\map f t}
      | c = {{Defof|Laplace Transform}}
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared by Root of a x + b

:$\ds \int x^2 \sqrt {a x + b} \rd x = \frac {2 \paren {15 a^2 x^2 - 12 a b x + 8 b^2} } {105 a^3} \sqrt {\paren {a x + b}^3} + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \sqrt {a x + b}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {u^2 - b} a
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int x \sqrt {a x + b} \rd x
      | r = \frac 2 a \int \paren {\frac {u^2 - b} a}^2 u^2 \rd u
      | c = Primitive of Function of $\sqrt {a x + b}$
}}
{{eqn | r = \frac 2 {a^3} \paren {\int u^6 \rd u - 2 b \int u^4 \rd u + b^2 \int u^2 \rd u}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 2 {a^3} \paren {\frac {u^7} 7 - 2 b \frac {u^5} 5 + b^2 \frac {u^3} 3} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac 2 {a^3} \paren {\frac {15 u^4 - 42 b u^2 + 35 b^2} {105} } u^3 + C
      | c = common denominator and $u^3$ as a factor
}}
{{eqn | r = \frac 2 {a^3} \paren {\frac {15 \paren {a x + b}^2 - 42 b \paren {a x + b} + 35 b^2} {105} } \sqrt {\paren {a x + b}^3} + C
      | c = substituting for $u$
}}
{{eqn | r = \frac {2 \paren {15 a^2 x^2 + 30 a b x + 15 b^2 - 42 a b x - 42 b^2 + 35 b^2} } {105 a^3} \sqrt {\paren {a x + b}^3} + C
      | c = multiplying out
}}
{{eqn | r = \frac {2 \paren {15 a^2 x^2 - 12 a b x + 8 b^2} } {105 a^3} \sqrt {\paren {a x + b}^3} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a x + b over x

:$\ds \int \frac {\sqrt {a x + b} } x \rd x = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }$

Proof:
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 $n = \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } x \rd x
      | r = \int x^{-1} \paren {a x + b}^{1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^0 \paren {a x + b}^{1/2} } {\frac 1 2} + \frac {\frac 1 2 b} {\frac 1 2} \int x^{-1} \paren {a x + b}^{- 1/2} \rd x
      | c = 
}}
{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a x + b over x

:$\ds \int \frac {\sqrt {a x + b} } x \rd x = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = v
      | r = \sqrt x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = \frac 1 {2 \sqrt x}
      | c = Power Rule for Derivatives
}}
{{eqn | l = u
      | r = \frac {2 \sqrt {a x + b} } {\sqrt x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {\frac {\sqrt x \cdot 2 a} {2 \sqrt{a x + b} } - \frac {2 \sqrt {a x + b} } {2 \sqrt x} } x
      | c = Quotient Rule for Derivatives etc.
}}
{{eqn | r = \frac {-b} {x^{3/2} \sqrt {a x + b} }
      | c = simplifying
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } x \rd x
      | r = \int \frac {2 \sqrt {a x + b} }{\sqrt x} \frac 1 {2 \sqrt x} \rd x
      | c = 
}}
{{eqn | r = \frac {2 \sqrt {a x + b} } {\sqrt x} \sqrt x - \int {\sqrt x} \frac {-b} {x^{3/2} \sqrt {a x + b} } \rd x
      | c = 
}}
{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a x + b over x squared

:$\ds \int \frac {\sqrt {a x + b} } {x^2} \rd x = -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }$

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 $n = \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } {x^2} \rd x
      | r = \int x^{-2} \paren {a x + b}^{1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^{-1} \paren {a x + b}^{1/2} } {-2 + \frac 1 2 + 1} + \frac {\frac 1 2 b} {-2 + \frac 1 2 + 1} \int x^{-2} \paren {a x + b}^{-1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {-2 \sqrt {a x + b} } x - b \int \frac {\d x} {x^2 \sqrt {a x + b} }
      | c = simplifying
}}
{{eqn | r = \frac {-2 \sqrt {a x + b} } x - b \paren {-\frac {\sqrt {a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt {a x + b} } }
      | c = Primitive of $\dfrac 1 {x^2 \sqrt {a x + b} }$
}}
{{eqn | r = \frac {-2 \sqrt {a x + b} } x + \frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }
      | c = multiplying out
}}
{{eqn | r = -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x over Root of a x + b

:$\ds \int \frac {x^m} {\sqrt {a x + b} } \rd x = \frac {2 x^m \sqrt{a x + b} } {\paren {2 m + 1} a} - \frac {2 m b} {\paren {2 m + 1} a} \int \frac {x^{m - 1} } {\sqrt {a x + b} } \rd x$

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Putting $n = -\dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {x^m} {\sqrt {a x + b} } \rd x
      | r = \int x^m \paren {a x + b}^{-1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^m \paren {a x + b}^{1/2} } {\paren {m - \frac 1 2 + 1} a} - \frac {m b} {\paren {m - \frac 1 2 + 1} a} \int x^{m - 1} \paren {a x + b}^{-1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^m \sqrt {a x + b} } {\paren {m + \frac 1 2} a} - \frac {m b} {\paren {m + \frac 1 2} a} \int \frac {x^{m - 1} } {\sqrt {a x + b} } \rd x
      | c = simplifying
}}
{{eqn | r = \frac {2 x^m \sqrt {a x + b} } {\paren {2 m + 1} a} - \frac {2 m b} {\paren {2 m + 1} a} \int \frac {x^{m - 1} } {\sqrt {a x + b} } \rd x
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of x by Root of a x + b

:$\ds \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$

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 $m := -m$:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^m \sqrt{a x + b} }
      | r = \int x^{-m} \paren {a x + b}^{-1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^{-m+1} \paren {a x + b}^{1/2} } {\paren {-m + 1} b} - \frac {\paren {-m - \frac 1 2 + 2} a} {\paren {-m + 1} b} \int x^{-m + 1} \paren {a x + b}^{-1/2} \rd x
      | c = 
}}
{{eqn | r = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {m - \frac 3 2} a} {\paren {m - 1} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }
      | c = simplifying
}}
{{eqn | r = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x by Root of a x + b

:$\ds \int x^m \sqrt {a x + b} \rd x = \frac {2 x^m} {\paren {2 m + 3} a} \paren {\sqrt {a x + b} }^3 - \frac {2 m b} {\paren {2 m + 3} a} \int x^{m - 1} \sqrt{a x + b} \rd x$

Proof:
From Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Putting $n := \dfrac 1 2$:
{{begin-eqn}}
{{eqn | l = \int x^m \sqrt{a x + b} \rd x
      | r = \int x^m \paren {a x + b}^{1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^m \paren {a x + b}^{3/2} } {\paren {m + \frac 1 2 + 1} a} - \frac {m b} {\paren {m + \frac 1 2 + 1} a} \int x^{m - 1} \paren {a x + b}^{1/2} \rd x
      | c = 
}}
{{eqn | r = \frac {x^m \paren {\sqrt {a x + b} }^3} {\paren {m + \frac 3 2} a} - \frac {m b} {\paren {m + \frac 3 2} a} \int x^{m - 1} \sqrt {a x + b} \rd x
      | c = simplifying
}}
{{eqn | r = \frac {2 x^m} {\paren {2 m + 3} a} \paren {\sqrt {a x + b} }^3 - \frac {2 m b} {\paren {2 m + 3} a} \int x^{m - 1} \sqrt {a x + b} \rd x
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Root of a x + b over Power of x/Formulation 1

:$\ds \int \frac {\sqrt {a x + b} } {x^m} \rd x = -\frac {\sqrt {a x + b} } {\paren {m - 1} x^{m - 1} } + \frac a {2 \paren {m - 1} } \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \sqrt {a x + b}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac a {2 \sqrt {a x + b} }
      | c = Power Rule for Derivatives etc.
}}
{{eqn | l = v
      | r = \frac {-1} {\paren {m - 1} x^{m - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^m}
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int \frac {\sqrt {a x + b} } {x^m} \rd x
      | r = \int \sqrt {a x + b} \frac 1 {x^m} \rd x
      | c = 
}}
{{eqn | r = \sqrt {a x + b} \frac {-1} {\paren {m - 1} x^{m - 1} } - \int \frac {-1} {\paren {m - 1} x^{m - 1} } \frac a {2 \sqrt {a x + b} } \rd x
      | c = 
}}
{{eqn | r = -\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} }
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
{{qed}}