hcju/proofwikiqa huggingface dataset:
Let: {{begin-eqn}} {{eqn | l = u       | r = \paren {a x + b}^m       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = m a \paren {a x + b}^m       | c = Power Rule for Derivatives and Derivatives of Function of $a x + b$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = v       | r = \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} }       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d v} {\d x}       | r = \paren {p x + q}^n       | c = Power Rule for Derivatives and Derivatives of Function of $a x + b$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | o =        | r = \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x       | c =  }} {{eqn | r = \paren {a x + b}^m \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } - \int \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } m a \paren {a x + b}^m \rd x       | c = Integration by Parts }} {{eqn | r = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m-1} } {\paren {p x + q}^{n - 1} } \rd x}       | c = simplification }} {{end-eqn}} {{qed}}
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 \frac {p x + q} {\sqrt {a x + b} } \rd x       | r = \int \paren {p \paren {\frac {u^2 - b} a} + q} \frac {2 u} {a u} \rd x       | c = Primitive of Function of Root of $a x + b$ }} {{eqn | r = \frac 2 {a^2} \int \paren {p u^2 - b p + a q} \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \frac 2 {a^2} \paren {\frac {p u^3} 3 - b p z + a q z} + C       | c = Primitive of Power }} {{eqn-intertext|Note that at this point we can assume $u > 0$, otherwise $\sqrt {a x + b}$ would not have been defined.}} {{eqn | r = \frac 2 {a^2} \paren {\frac {p \paren {\sqrt {a x + b} }^3} 3 - b p \sqrt {a x + b} + a q \sqrt {a x + b} } + C       | c = substituting for $u$ }} {{eqn | r = \frac {2 \paren {a p x - 2 b p + 3 a q} } {3 a^2} \sqrt {a x + b} + C       | c = simplification }} {{end-eqn}} {{qed}}
{{finish|Please deal with positive and negative}} 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 \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }       | r = \int \frac {2 u} {p \paren {\paren {\dfrac {u^2 - b} a} + q} a u} \rd u       | c = Primitive of Function of $\sqrt{a x + b}$ }} {{eqn | r = \int \frac {2 a u} {\paren {p \paren {u^2 - b} + a q} a u} \rd u       | c = simplifying }} {{eqn | r = 2 \int \frac {\d u} {p u^2 - b p + a q}       | c = simplifying }} {{eqn | n = 1       | r = 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}       | c = putting denominator into the form $z^2 - a^2$ }} {{end-eqn}} Now let $z = \sqrt p u$ {{begin-eqn}} {{eqn | l = z       | r = \sqrt p u       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d u}       | r = \sqrt p       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }       | r = 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}       | c = from $(1)$ }} {{eqn | n = 2       | r = \frac 2 {\sqrt p} \int \frac {\d z} {z^2 - \paren {\sqrt {b p - a q} }^2}       | c = simplifying }} {{end-eqn}} Let $b p - a q > 0$. Let $d = \sqrt {b p - a q}$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }       | r = \frac 2 {\sqrt p} \int \frac {\d z} {z^2 - d^2}       | c = from $(2)$ }} {{eqn | r = \frac 2 {\sqrt p} \frac 1 {2 d} \ln \size {\frac {z - d} {z + d} } + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form }} {{eqn | r = \frac 1 {\sqrt p d} \ln \size {\frac {\sqrt p u - d} {\sqrt p u + d} } + C       | c = substituting for $z$ }} {{eqn | r = \frac 1 {\sqrt p \sqrt {b p - a q} } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C       | c = substituting for $u$ and $d$ }} {{end-eqn}} {{qed|lemma}} Let $b p - a q < 0$. Let $d = \sqrt {-\paren {b p - a q} } = \sqrt {a q - b p}$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }       | r = \frac 2 {\sqrt p} \int \frac {\d z} {z^2 + d^2}       | c = from $(2)$ }} {{eqn | r = \frac 2 {\sqrt p} \frac 1 d \arctan {\frac z d} + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac 2 {\sqrt p d} \arctan {\frac {\sqrt p u} d} + C       | c = substituting for $z$ }} {{eqn | r = \frac 2 {\sqrt {a q - b p} \sqrt p} \arctan {\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C       | c = substituting for $u$ and $d$ }} {{end-eqn}} {{qed}}
From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 2: :$\displaystyle \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$ Setting $m := \dfrac 1 2$ and $n = 1$: {{begin-eqn}} {{eqn | l = \int \frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^n} \rd x       | r = \frac {-1} {\paren {1 - \frac 1 2 - 1} p} \paren {\frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^0} + \frac 1 2 \paren {b p - a q} \int \frac {\paren {a x + b}^{1/2 - 1} } {\paren {p x + q} } \rd x}       | c =  }} {{eqn | r = \frac {2 \sqrt {a x + b} } p + \frac {b p - a q} p \int \frac {\d x} {\sqrt {a x + b} \paren {p x + q} }       | c =  }} {{end-eqn}} From Primitive of Reciprocal of $p x + q$ by Root of $a x + b$: :$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} } = \begin{cases} \dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\ \dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\ \end{cases}$ The result follows by substitution. {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power: :$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$ Setting $m := \dfrac 1 2$: {{begin-eqn}} {{eqn | l = \int \paren {a x + b}^{1 / 2} \paren {p x + q}^n \rd x       | r = \frac {\paren {a x + b}^{1 / 2} \paren {p x + q}^{n + 1} } {\paren {\frac 1 2 + n + 1} p} + \frac {\dfrac 1 2 \paren {b p - a q} } {\paren {\frac 1 2 + n + 1} p} \int \paren {a x + b}^{1 / 2 - 1} \paren {p x + q}^n \rd x       | c =  }} {{eqn | r = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x       | c =  }} {{end-eqn}} {{qed}}
From Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)^m \left({p x + q}\right)^n}$: :$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$ Setting $m := \dfrac 1 2$: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} }       | r = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{1/2 - 1} \left({p x + q}\right)^{n-1} } + a \left({\frac 1 2 + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^{1/2} \left({p x + q}\right)^{n-1} } }\right)       | c =  }} {{eqn | r = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right)  \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }       | c =  }} {{end-eqn}} {{qed}}
From Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power: :$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$ Setting $m := -\dfrac 1 2$: {{begin-eqn}} {{eqn | l = \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x       | r = \frac {\left({a x + b}\right)^{-1/2 + 1} \left({p x + q}\right)^n} {\left({-\frac 1 2 + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({-\frac 1 2 + n + 1}\right) a} \int \left({a x + b}\right)^{-1/2} \left({p x + q}\right)^{n-1} \ \mathrm d x       | c =  }} {{eqn | r = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x       | c =  }} {{end-eqn}} {{qed}}
From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 3: :$\displaystyle \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n} \ \mathrm d x = \frac {-1} {\left({n - 1}\right) p} \left({\frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^{n-1} } + m a \int \frac {\left({a x + b}\right)^{m-1} } { \left({p x + q}\right)^{n-1} } \ \mathrm d x}\right)$ Setting $m := \dfrac 1 2$: {{begin-eqn}} {{eqn | l = \int \frac {\sqrt{a x + b} } {\left({p x + q}\right)^n} \ \mathrm d x       | r = \frac {-1} {\left({n - 1}\right) p} \left({\frac {\left({a x + b}\right)^{1/2} } {\left({p x + q}\right)^{n-1} } + \frac 1 2 a \int \frac {\left({a x + b}\right)^{1/2-1} } { \left({p x + q}\right)^{n-1} } \ \mathrm d x}\right)       | c =  }} {{eqn | r =\frac {- \sqrt{a x + b} } {\left({n - 1}\right) p \left({p x + q}\right)^{n-1} } + \frac a {2 \left({n - 1}\right) p} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = u       | r = \sqrt {a x + b}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac {u^2 - b} a       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {p x + q}       | r = \sqrt {p \paren {\frac {u^2 - b} a} + q}       | c =  }} {{eqn | r = \sqrt {\frac {p \paren {u^2 - b} + a q} a}       | c =  }} {{eqn | r = \sqrt {\frac {p u^2 - b p + a q} a}       | c =  }} {{eqn | n = 1       | r = \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} }       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }       | r = \int \frac {2 u \rd u} {a \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} } u}       | c = Primitive of Function of Root of $a x + b$ }} {{eqn | r = \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } }       | c = Primitive of Constant Multiple of Function }} {{eqn-intertext|Setting $c^2 :{{=}} \dfrac {b p - a q} p$:}} {{eqn | n = 2       | r = \frac 2 {\sqrt {a p} } \int \frac {\rd u} {\sqrt{u^2 - c^2} }       | c =  }} {{end-eqn}} Let $c^2 > 0$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }       | r = \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt{u^2 - c^2} }       | c = from $(2)$ }} {{eqn | r = \frac 2 {\sqrt {a p} } \ln \size {u + \sqrt {u^2 - c^2} } + C       | c = Primitive of Reciprocal of $\sqrt {x^2 - a^2}$: Logarithm Form }} {{eqn | r = \frac 2 {\sqrt {a p} } \ln \size {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C       | c = substituting for $u$ and $c$ }} {{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C       | c = as both those square roots are positive }} {{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {a x + b} + \sqrt {\frac a p} \sqrt {p x + q} } + C       | c = substituting from $(1)$ }} {{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\frac {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } {\sqrt p} } + C       | c = simplifying }} {{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } - \ln {\sqrt p} + C       | c = Difference of Logarithms }} {{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C       | c = subsuming $-\ln {\sqrt p}$ into the arbitrary constant }} {{end-eqn}} {{qed|lemma}} Let $c^2 < 0$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }       | r = \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 + c^2} }       | c = from $(2)$ }} {{eqn | r = \frac 2 {\sqrt {a p} } \sinh^{-1} {\frac u c} + C       | c = Primitive of Reciprocal of $\sqrt {x^2 + a^2}$: $\sinh^{-1}$ form }} {{eqn | r = \frac 2 {\sqrt {a p} } \sinh^{-1} \frac {\sqrt {a x + b} } {\sqrt {\frac {b p - a q} p} } + C       | c = substituting for $u$ and $c$ }} {{eqn | r = \frac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\frac {p \paren {a x + b} } {b p - a q} } + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{mistake|see talk page}} {{begin-eqn}} {{eqn | l = \frac \d {\d s} \laptrans {\map f t}       | r = \frac \d {\d s} \int_0^{\to +\infty} \map f t \, e^{-s t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^{\to +\infty} \map {\frac {\partial} {\partial s} } {\map f t \, e^{-s t} } \rd t       | c = Definite Integral of Partial Derivative }} {{eqn | r = \int_0^{\to +\infty} \map f t \, \map {\frac {\partial} {\partial s} } {e^{-st} } \rd t }} {{eqn | r = -\int_0^{\to +\infty} t \, \map f t \, e^{-s t} \rd t       | c = Derivative of Exponential Function }} {{eqn | r = -\laptrans {t \, \map f t}       | c = {{Defof|Laplace Transform}} }} {{end-eqn}} {{qed}}
The proof proceeds by induction on $n$, the order of the derivative of $\laptrans f$. The case $n = 0$ is verified as follows: {{begin-eqn}} {{eqn | l = \frac {\d^0} {\d s^0} \laptrans {\map f t}       | r = \laptrans {\map f t}       | c = {{Defof|Zeroth Derivative}} }} {{eqn | r = \paren {-1}^0 \laptrans {t^0 \, \map f t}       | c = {{Defof|Power of Natural Number|Zeroth Power}} }} {{end-eqn}} === Basis for the Induction === The case $n = 1$ is demonstrated in Derivative of Laplace Transform: :$\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$ This is the basis for the induction. === Induction Hypothesis === Fix $n \in \N$ with $n \ge 0$. Assume: :$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$ This is our induction hypothesis. === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \frac {\d^{n + 1} } {\d s^{n + 1} } \laptrans {\map f t}       | r = \frac {\d}{\d s} \frac {\d^n} {\d s^n} \laptrans {\map f t} }} {{eqn | r = \frac {\d}{\d s} \paren {\paren {-1}^n \laptrans {t^n \, \map f t} }       | c = Induction Hypothesis }} {{eqn | r = \paren {-1}^n \paren {-\laptrans {t \times t^n \, \map f t} }       | c = Derivative of Laplace Transform }} {{eqn | r = \paren {-1}^{n + 1} \laptrans {t^{n + 1} \, \map f t}       | c = simplification }} {{end-eqn}} The result follows by the Principle of Mathematical Induction. {{qed}}
From Primitive of $\paren {p x + q}^n \sqrt {a x + b}$: :$\displaystyle \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x$ Putting $n = \dfrac 1 2$: {{begin-eqn}} {{eqn | o =        | r = \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x       | c =  }} {{eqn | r = \frac {2 \paren {p x + q}^{1 / 2 + 1} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 3} p} + \frac {b p - a q} {\paren {2 \cdot \frac 1 2 + 3} p} \int \frac {\paren {p x + q}^{1 / 2} } {\sqrt {a x + b} } \rd x       | c =  }} {{eqn | r = \frac {2 \paren {p x + q} \sqrt {p x + q} \sqrt {a x + b} } {4 p} + \frac {b p - a q} {4 p} \int \frac {\sqrt {p x + q} } {\sqrt {a x + b} } \rd x       | c =  }} {{eqn | r = \frac {2 \paren {p x + q} \sqrt {p x + q} \sqrt {a x + b} } {4 p} + \frac {b p - a q} {4 p} \paren {\frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } }       | c = Primitive of $\dfrac {\sqrt {p x + q} } {\sqrt {a x + b} }$ }} {{eqn | r = \frac {\paren {2 a \paren {p x + q} + \paren {b p - a q} } \sqrt {\paren {a x + b} \paren {p x + q} } } {4 a p} - \frac {\paren {b p - a q}^2 } {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }       | c = extracting common factor }} {{eqn | r = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }       | c = simplifying }} {{end-eqn}} {{qed}}
From Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$: :$\displaystyle \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$ Putting $n = \dfrac 1 2$: {{begin-eqn}} {{eqn | l = \int \sqrt {\frac {p x + q} {a x + b} } \rd x       | r = \frac {2 \paren {p x + q}^{1/2} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 1} a} + \frac {2 \cdot \frac 1 2 \paren {a q - b p} } {\paren {2 \cdot \frac 1 2 + 1} a} \int \frac {\paren {p x + q}^{1/2 - 1} } {\sqrt {a x + b} } \rd x       | c =  }} {{eqn | r = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }       | c =  }} {{end-eqn}} {{qed}}
From Primitive of $\paren {p x + q}^n \sqrt {a x + b}$: :$\displaystyle \int \frac {\d x} {\paren {p x + q}^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} }$ Putting $n = \dfrac 3 2$: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }       | r = \frac {\sqrt {a x + b} } {\paren {\frac 3 2 - 1} \paren {a q - b p} \paren {p x + q}^{3 / 2 - 1} } + \frac {\paren {2 \cdot \frac 3 2 - 3} a} {2 \paren {\frac 3 2 - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{3 / 2 - 1} \sqrt {a x + b} }       | c =  }} {{eqn | r = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + \frac {\paren {3 - 3} a} {a q - b p} \int \frac {\d x} {\sqrt {p x + q} \sqrt {a x + b} } + C       | c =  }} {{eqn | r = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = u       | r = x^2 + a^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = 2 x        | c = Power Rule for Derivatives and Derivative of Constant }} {{eqn | ll= \leadsto       | l = \int \frac {x \rd x} {x^2 + a^2}       | r = \frac 1 2 \ln \size {x^2 + a^2} + C       | c = Primitive of Function under its Derivative }} {{eqn | r = \frac 1 2 \, \map \ln {x^2 + a^2} + C       | c = Absolute Value of Even Power }} {{end-eqn}} {{qed}}
From Primitive of Power of x less one over Power of x plus Power of a: :$\displaystyle \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$ So: {{begin-eqn}} {{eqn | l = \int \frac {x \rd x} {x^2 + a^2}       | r = \frac 1 2 \ln \size {x^2 + a^2} + C       | c = Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 2$ }} {{eqn | r = \frac 1 2 \, \map \ln {x^2 + a^2} + C       | c = Absolute Value of Even Power }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \rd x} {x^2 + a^2}       | r = \int \paren {1 - \frac {a^2} {x^2 + a^2} } \rd x       | c = long division }} {{eqn | r = \int \d x - a^2 \int \frac {\d x} {x^2 + a^2}        | c = Linear Combination of Integrals }} {{eqn | r = x - a^2 \int \frac {\d x} {x^2 + a^2} + C       | c = Primitive of Constant }} {{eqn | r = x - a^2 \frac 1 a \arctan {\frac x a} + C       | c = Primitive of Reciprocal of $x^2 + a^2$ }} {{eqn | r = x - a \arctan {\frac x a} + C       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = x       | r = a \tan \theta       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d \theta}       | r = a \sec^2 \theta       | c = Derivative of Tangent Function }} {{eqn | ll= \leadsto       | l = \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}       | r = \int \frac {a^2 \tan^2 \theta a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta       | c = Integration by Substitution }} {{eqn | r = \int \frac {a^3 \tan^2 \theta \sec^2 \theta} {a^4 \sec^4 \theta} \rd \theta       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac 1 a \int \frac {\tan^2 \theta} {\sec^2 \theta} \rd \theta       | c = simplification }} {{eqn | r = \frac 1 a \int \tan^2 \theta \cos^2 \theta \rd \theta       | c = {{Defof|Real Secant Function}} }} {{eqn | r = \frac 1 a \int \frac {\sin^2 \theta} {\cos^2 \theta} \cos^2 \theta \rd \theta       | c = {{Defof|Real Tangent Function}} }} {{eqn | r = \frac 1 a \int \sin^2 \theta \rd \theta       | c = simplification }} {{eqn | r = \frac \theta {2 a} - \frac {\sin 2 \theta} {4 a} + C       | c = Primitive of Square of Sine Function }} {{eqn | r = \frac \theta {2 a}  - \frac 1 {2 a} \frac {\tan \theta} {1 + \tan^2 \theta} + C       | c = Tangent Half-Angle Substitution for Sine }} {{eqn | r = -\frac x {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C       | c = substituting for $\theta$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}       | r = \int \frac {x^2 + a^2 - a^2} {\paren {x^2 + a^2}^2} \rd x       | c =  }} {{eqn | r = \int \frac {x^2 + a^2} {\paren {x^2 + a^2}^2} \rd x - a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {\d x} {x^2 + a^2} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}       | c = simplification }} {{eqn | r = \frac 1 a \arctan {\frac x a} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2} + C       | c = Primitive of Reciprocal of $x^2 + a^2$ }} {{eqn | r = \frac 1 a \arctan {\frac x a} + a^2 \paren {\frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a} + C       | c = Primitive of Reciprocal of $\paren {x^2 + a^2}^2$ }} {{eqn | r = \frac {-x} {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^3 \rd x} {x^2 + a^2}       | r = \int \paren {x - \frac {a^2 x} {x^2 + a^2} } \rd x       | c = long division }} {{eqn | r = \int x \rd x - a^2 \int \frac {x \rd x} {x^2 + a^2}        | c = Linear Combination of Integrals }} {{eqn | r = \frac {x^2} 2 - a^2 \int \frac {x \rd x} {x^2 + a^2} + C       | c = Primitive of Power }} {{eqn | r = \frac {x^2} 2 - a^2 \paren {\frac 1 2 \map \ln {x^2 + a^2} } + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | r = \frac {x^2} 2 - \frac {a^2} 2 \map \ln {x^2 + a^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}