hcju/proofwikiqa huggingface dataset:
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}} Thus: {{begin-eqn}} {{eqn | l = \map F {\sqrt {a x + b} }       | r = \frac 1 {x \sqrt {a x + b} }       | c =  }} {{eqn | ll= \leadsto       | l = \map F u       | r = \paren {\frac a {u^2 - b} } \frac 1 u       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \sqrt {a x + b} }       | r = \frac 2 a \int u \paren {\frac a {u^2 - b} } \frac 1 u \rd u       | c = Primitive of Function of $\sqrt {a x + b}$ }} {{eqn | r = 2 \int \frac {\d u} {u^2 - b}       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} Let $b > 0$. Let $d = \sqrt b$. Then: {{begin-eqn}} {{eqn | l = 2 \int \frac {\d u} {u^2 - b}       | r = 2 \int \frac {\d u} {u^2 - d^2}       | c =  }} {{eqn | r = 2 \paren {\frac 1 {2 d} } \map \ln {\frac {u - d} {u + d} } + C       | c = Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form }} {{eqn | r = \dfrac 1 {\sqrt b} \map \ln {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C       | c = substituting for $u$ and $d$ }} {{end-eqn}} {{qed|lemma}} {{questionable|In the link to the above primitive evaluation, the ln is of the absolute value of the argument. If $u - d$ can possibly be negative, this needs to hold over to this page If $x$ is negative, then this could indeed be the case. Hence the source work may also need to be scrutinised to see whether it has glossed over this detail.}} Let $b < 0$. Let $d = \sqrt {-b}$. Then: {{begin-eqn}} {{eqn | l = 2 \int \frac {\d u} {u^2 - b}       | r = 2 \int \frac {\d u} {u^2 - \paren {-d^2} }       | c =  }} {{eqn | r = 2 \int \frac {\d u} {u^2 + d^2}       | c =  }} {{eqn | r = 2 \paren {\frac 1 d} \arctan {\frac u d} + C       | c = Primitive of Reciprocal of $x^2 + a^2$ }} {{eqn | r = \frac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C       | c = substituting for $u$ and $d$ }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$: :$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$ Setting $m = -2$ and $n = -\dfrac 1 2$: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \sqrt{a x + b} }       | r = \int x^{-2} \left({a x + b}\right)^{-1/2} \rd x       | c =  }} {{eqn | r = \frac {x^{-1} \left({a x + b}\right)^{1/2} } {\left({-1}\right) b} - \frac {\left({-1/2}\right) a} {\left({-1}\right) b} \int x^{-1} \left({a x + b}\right)^{-1/2} \rd x       | c =  }} {{eqn | r = -\frac {\sqrt{a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt{a x + b} }       | c =  }} {{end-eqn}} {{qed}}
Let $u = \sqrt{a x + b}$. Then: {{begin-eqn}} {{eqn | l = \int \sqrt{a x + b} \ \mathrm d x       | r = \frac 2 a \int u^2 \ \mathrm d u       | c = Primitive of Function of $\sqrt{a x + b}$ }} {{eqn | r = \frac 2 a \frac {u^3} 3       | c = Primitive of Power }} {{eqn | r = \frac {2 \sqrt{\left({a x + b}\right)^3} } {3 a}       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$: :$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$ 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}}
Let: {{begin-eqn}} {{eqn | l = v       | r = \sqrt x       | c =  }} {{eqn | ll= \implies       | 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= \implies       | 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: : $\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$ 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}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \sinh^{-1} {\frac x a}       | c =  }} {{eqn | l = x       | r = a \sinh u       | c =  }} {{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 = \sinh^{-1} {\frac x a} + C       | c = Definition of $u$ }} {{end-eqn}} {{qed}}
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 {\cosh^{-1} } {\frac x a}       | c =  }} {{eqn | 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 {\cosh^{-1} } {\frac x a} + C       | c = Definition of $u$ }} {{eqn | r = \cosh^{-1} \size {\frac x a} + C       | c = as $\dfrac x a > 0$ }} {{eqn | r = \dfrac {\size x} x \cosh^{-1} \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 {\cosh^{-1} } {\frac z a} + C       | c = from above }} {{eqn | r = -\map {\cosh^{-1} } {\frac {-x} a} + C       | c = substituting back }} {{eqn | r = -\cosh^{-1} {\size {\frac x a} } + C       | c = as $\dfrac x a < 0$ }} {{eqn | r = \dfrac {\size x} x \cosh^{-1} {\size {\frac x a} } + C       | c = as $x < 0$, so $\dfrac {\size x} x = -1$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \csch^{-1} {\frac x a}       | c =  }} {{eqn | 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}} {{finish|There may be a complication with the sign of the argument which needs to be resolved}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \sech^{-1} {\frac x a}       | c =  }} {{eqn | 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}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \sqrt{a x + b}       | c =  }} {{eqn | ll= \implies       | l = x       | r = \frac {u^2 - b} a       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \sqrt{a x + b} \ \mathrm d x       | r = \frac 2 a \int \frac {u^2 - b} a u^2 \ \mathrm d u       | c = Primitive of Function of $\sqrt{a x + b}$ }} {{eqn | r = \frac 2 {a^2} \int u^4 \ \mathrm d u - \frac {2 b} {a^2} \int u^2 \ \mathrm d u       | c = Linear Combination of Integrals }} {{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 {\left({6 u^2 - 10 b}\right) u^3} {15 a^2} + C       | c = extracting $u^3$ as a factor }} {{eqn | r = \frac {6 \left({a x + b}\right) - 10 b} {15 a^2} \sqrt{\left({a x + b}\right)^3} + C       | c = substituting for $u$ }} {{eqn | r = \frac {2 \left({3 a x - 2 b}\right)  } {15 a^2} \sqrt{\left({a x + b}\right)^3} + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{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}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \sqrt{a x + b}       | c =  }} {{eqn | ll= \implies       | l = x       | r = \frac {u^2 - b} a       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int x \sqrt{a x + b} \ \mathrm d x       | r = \frac 2 a \int \left({\frac {u^2 - b} a}\right)^2 u^2 \ \mathrm d u       | c = Primitive of Function of $\sqrt{a x + b}$ }} {{eqn | r = \frac 2 {a^3} \left({\int u^6 \ \mathrm d u - 2 b \int u^4 \ \mathrm d u + b^2 \int u^2 \ \mathrm d u}\right)       | c = Linear Combination of Integrals }} {{eqn | r = \frac 2 {a^3} \left({\frac {u^7} 7 - 2 b \frac {u^5} 5 + b^2 \frac {u^3} 3}\right) + C       | c = Primitive of Power }} {{eqn | r = \frac 2 {a^3} \left({\frac {15 u^4 - 42 b u^2 + 35 b^2} {105} }\right) u^3 + C       | c = common denominator and $u^3$ as a factor }} {{eqn | r = \frac 2 {a^3} \left({\frac {15 \left({a x + b}\right)^2 - 42 b \left({a x + b}\right) + 35 b^2} {105} }\right) \sqrt{\left({a x + b}\right)^3} + C       | c = substituting for $u$ }} {{eqn | r = \frac {2 \left({15 a^2 x^2 + 30 a b x + 15 b^2 - 42 a b x - 42 b^2 + 35 b^2}\right)} {105 a^3} \sqrt{\left({a x + b}\right)^3} + C       | c = multiplying out }} {{eqn | r = \frac {2 \left({15 a^2 x^2 - 12 a b x + 8 b^2}\right)} {105 a^3} \sqrt{\left({a x + b}\right)^3} + C       | c = simplifying }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$: :$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$ 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 Reciprocal of x squared by Root of 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}}
From Reduction Formula for 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}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$: :$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$ 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}}
Let $u = \sqrt{a x + b}$. Then: {{begin-eqn}} {{eqn | l = \int \left({\sqrt{a x + b} }\right)^m \ \mathrm d x       | r = \frac 2 a \int u^{m + 1} \ \mathrm d x       | c = Primitive of Function of $a x + b$ }} {{eqn | r = \frac 2 a \frac {u^{m + 2} } {m + 2} + C       | c = Primitive of Power }} {{eqn | r = \frac {2 \left({\sqrt{a x + b} }\right)^{m + 2} } {a \left({m + 2}\right)} + C       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$: :$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \ \mathrm d x$ Putting $n := \dfrac m 2$ and $m := -1$: {{begin-eqn}} {{eqn | l = \int \frac {\left({\sqrt{a x + b} }\right)^m} x \ \mathrm d x       | r = \int x^{-1} \left({a x + b}\right)^{m/2} \ \mathrm d x       | c =  }} {{eqn | r = \frac {x^0 \left({a x + b}\right)^{m/2} } {-1 + \frac m 2 + 1} + \frac {\left({\frac m 2}\right) b} {-1 + \frac m 2 + 1} \int x^{-1} \left({a x + b}\right)^{m/2 - 1} \ \mathrm d x       | c =  }} {{eqn | r = \frac {2 \left({a x + b}\right)^{m/2} } m + b \int x^{-1} \left({a x + b}\right)^{\left({m - 2}\right)/2} \ \mathrm d x       | c = simplifying }} {{eqn | r = \frac {2 \left({\sqrt{a x + b} }\right)^m } m + b \int \frac {\left({\sqrt{a x + b} }\right)^{m - 2} } x \ \mathrm d x       | c = }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $a x + b$: :$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {-x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) b} + \frac {m + n + 2} {\left({n + 1}\right) b} \int x^m \left({a x + b}\right)^{n + 1} \ \mathrm d x$ Putting $n := -\dfrac m 2$ and $m := -1$: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^m}       | r = \int x^{-1} \left({a x + b}\right)^{-m/2}       | c =  }} {{eqn | r = \frac {-x^0 \left({a x + b}\right)^{-m/2 + 1} } {\left({- \frac m 2 + 1}\right) b} + \frac {-1 - \frac m 2 + 2} {\left({- \frac m 2 + 1}\right) b} \int x^{-1} \left({a x + b}\right)^{- m/2 + 1} \ \mathrm d x       | c =  }} {{eqn | r = \frac 2 {\left({m - 2}\right) b \left({a x + b}\right)^{\left({m - 2}\right)/2} } - \frac 1 b \int \frac {\mathrm d x} {x \left({a x + b}\right)^{\left({m - 2}\right) / 2} }       | c = simplifying }} {{eqn | r = \frac 2 {\left({m - 2}\right) b \left({\sqrt{a x + b} }\right)^{m - 2} }  + \frac 1 b \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^{m - 2} }       | c = }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)}       | r = \int \left({\frac {-a} {\left({b p - a q}\right) \left({a x + b}\right)} + \frac p {\left({b p - a q}\right) \left({p x + q}\right)} }\right)\ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {b p - a q} \left({ {-a} \int \frac 1 {a x + b} \ \mathrm d x + p \int \frac 1 {p x + q} \ \mathrm d x}\right)       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {b p - a q} \left({\frac {-a} a \ln \left\vert{a x + b}\right\vert + \frac p p \ln \left\vert{p x + q}\right\vert}\right) + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | r = \frac 1 {b p - a q} \left({\ln \left\vert{p x + q}\right\vert - \ln \left\vert{a x + b}\right\vert}\right) + C       | c = simplification }} {{eqn | r = \frac 1 {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)}       | r = \int \left({\frac b {\left({b p - a q}\right) \left({a x + b}\right)} - \frac q {\left({b p - a q}\right) \left({p x + q}\right)} }\right)\ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {b p - a q} \left({b \int \frac 1 {a x + b} \ \mathrm d x - q \int \frac 1 {p x + q} \ \mathrm d x}\right)       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {b p - a q} \left({\frac b a \ln \left\vert{a x + b}\right\vert - \frac q p \ln \left\vert{p x + q}\right\vert}\right) + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \int \frac {\mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)}       | c =  }} {{eqn | r = \int \left({\frac 1 {b p - a q} \left({\frac {-a p} {\left({b p - a q}\right) \left({a x + b}\right)} + \frac {-a} {\left({a x + b}\right)^2} + \frac {p^2} {\left({b p - a q}\right) \left({p x + q}\right)} }\right)}\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {b p - a q} \left({\frac {-a p} {b p - a q} \int \frac {\mathrm d x} {a x + b} - a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {p^2} {b p - a q} \int \frac {\mathrm d x} {p x + q} }\right)       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {b p - a q} \left({\frac {-a p} {b p - a q} \frac 1 a \ln \left\vert{a x + b}\right\vert - a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {p^2} {b p - a q} \frac 1 p \ln \left\vert{p x + q}\right\vert }\right) + C       | c = Primitive of Reciprocal of $a x + b$ }} {{eqn | r = \frac 1 {b p - a q} \left({\frac {-p} {b p - a q} \ln \left\vert{a x + b}\right\vert - a \frac {-1} {a \left({a x + b}\right)} + \frac p {b p - a q} \ln \left\vert{p x + q}\right\vert }\right) + C       | c = Primitive of Reciprocal of $\left({a x + b}\right)^2$ }} {{eqn | r = \frac 1 {b p - a q} \left({\frac 1 {a x + b} + \frac p {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert}\right) + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}