hcju/proofwikiqa huggingface dataset:
Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \bigcup_{j \mathop = 1}^n A_j \in \mathcal R$ $P(1)$ is true, as this just says $A_1 \in \mathcal R$. === Basis for the Induction === $P(2)$ is the case: : $A_1 \cup A_2 \in \mathcal R$ which is immediate, from definition 2 of ring of sets. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \bigcup_{j \mathop = 1}^k A_j \in \mathcal R$ Then we need to show: :$\displaystyle \bigcup_{j \mathop = 1}^{k+1} A_j \in \mathcal R$ === Induction Step === This is our induction step: We have that: :$\displaystyle \bigcup_{j \mathop = 1}^{k+1} A_j = \bigcup_{j \mathop = 1}^k A_j \cup A_{k+1}$ But from the induction hypothesis we have that: :$\displaystyle \bigcup_{j \mathop = 1}^k A_j \in \mathcal R$ Hence from the basis for the induction, it follows that: :$\displaystyle \bigcup_{j \mathop = 1}^{k+1} A_j \in \mathcal R$ So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \bigcup_{j \mathop = 1}^n A_j \in \mathcal R$ {{qed}}
From Gamma Difference Equation: :$\map \Gamma {z + 1} = z \, \map \Gamma z$ which is valid for all $z \notin \Z_{\le 0}$. The result follows by dividing by $z$.
{{Proofread}} From the Weierstrass form of the Gamma function: :$\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \left({\left({1 + \frac z n}\right) e^{-z / n} }\right)$ Taking the reciprocal of both sides: :$\displaystyle \Gamma \left({z}\right) = \frac {e^{-\gamma z}} z \prod_{n \mathop = 1}^\infty \frac {e^{z/n}} {1 + \frac z n}$ Taking the derivative of both sides: {{begin-eqn}} {{eqn | l = \Gamma\,' \left({z}\right)       | r = -\frac {e^{-\gamma z} \left({1 + \gamma z}\right)} {z^2} \prod_{n \mathop = 1}^\infty \left({\frac {e^{z / n} } {\left({1 + \frac z n}\right)} }\right) + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \left({\frac z {n \left({z + n}\right)} \prod_{i \mathop = 1}^\infty \frac {e^{z / i} } {1 + \frac z i} }\right)       | c = General Product Rule for Derivatives }} {{eqn | r = -\frac {e^{-\gamma z} \left({1 + \gamma z}\right)} {z^2} \frac z {e^{-\gamma z} } \Gamma \left({z}\right) + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \left({\frac z {n \left({z + n}\right)} \frac z {e^{-\gamma z} } \Gamma \left({z} \right) }\right)       | c = simplifying the product notation }} {{eqn | r = -\frac{1 + \gamma z} z \Gamma \left({z}\right) + \sum_{n \mathop = 1}^\infty \left({\frac {z \Gamma \left({z} \right)} {n \left({z + n}\right)} }\right)       | c = simplifying }} {{end-eqn}} Dividing both sides by $\Gamma \left({z}\right)$: {{begin-eqn}} {{eqn | l = \frac {\Gamma\,' \left({z}\right)} {\Gamma \left({z} \right)}       | r = -\frac {1 + \gamma z} z + \sum_{n \mathop = 1}^\infty \left({ \frac z {n \left({z + n}\right)} }\right) }} {{eqn | r = -\gamma - \frac 1 z + \sum_{n \mathop = 1}^\infty \left({ \frac 1 n - \frac 1 {z + n} }\right) }} {{eqn | r = -\gamma + \sum_{n \mathop = 1}^\infty \left({ \frac 1 n - \frac 1 {z + n - 1} }\right)       | c = rearranging the series }} {{end-eqn}} {{qed}}
From Reciprocal times Derivative of Gamma Function: :$\displaystyle \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$ Setting $n = 1$: {{begin-eqn}} {{eqn | l = \frac {\map {\Gamma'} 1} {\map \Gamma 1}       | r = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {1 + n - 1} }       | c =  }} {{eqn | r = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 n}       | c =  }} {{eqn | r = -\gamma + 0       | c =  }} {{eqn | r = -\gamma       | c =  }} {{end-eqn}} Using Gamma Function Extends Factorial: :$\map \Gamma 1 = \paren {1 - 1}! = 1$ Hence: :$\map {\Gamma'} 1 = -\gamma \map \Gamma 1 = -\gamma$ {{qed}}
{{proof wanted|Use Binet's Formula for Logarithm of Gamma Function/Formulation 2.}}
{{begin-eqn}} {{eqn | l = \map \Beta {x, y}       | r = \frac {\map \Gamma x \map \Gamma y} {\map \Gamma {x + y} }       | c = {{Defof|Beta Function|index = 3}} }} {{eqn | r = \frac {\map \Gamma y \map \Gamma x} {\map \Gamma {y + x} }       | c = Commutative Law of Addition and Commutative Law of Multiplication }} {{eqn | r = \map \Beta {y, x}       | c = {{Defof|Beta Function|index = 3}} }} {{end-eqn}} {{qed}}
Consider the substitution $s = \dfrac t {1 + t}$. We have the following: :$\dfrac 1 {1 + t} = 1 - s$ :$t \to 0, s \to 0$ :$t \to \infty, s \to 1$ :$\rd s = \dfrac 1 {\paren {1 + t}^2} \rd t$ Then: {{begin-eqn}} {{eqn | ll= \leadsto       | l = \int_{\to 0}^{\to \infty} \frac {t^{x - 1} } {\paren {1 + t}^{x + y} } \rd t       | r = \int_{\to 0}^{\to \infty} \paren {\frac t {1 + t} }^{x - 1} \paren {\frac 1 {1 + t} }^{y - 1} \frac {\rd t} {\paren {1 + t}^2}       | c =  }} {{eqn | r = \int_{\to 0}^{\to 1} s^{x - 1} \paren {1 - s}^{y - 1} \rd s       | c = Integration by Substitution }} {{eqn | r = \map \Beta {x, y}       | c = {{Defof|Beta Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {g_1} x \map {h_1} y + \map {g_2} x \map {h_2} y \frac {\d y} {\d x}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x}       | r = 0       | c = dividing both sides by $\map {g_2} x \map {h_1} y$ }} {{eqn | ll= \leadsto       | l = \int \left({\frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} }\right) \rd x       | r = \int 0 \rd x       | c = integrating both sides {{WRT|Integration}} $x$ }} {{eqn | ll= \leadsto       | l = \int \left({\frac {\map {g_1} x} {\map {g_2} x} + \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} }\right) \rd x       | r = C       | c = Primitive of Constant }} {{eqn | ll= \leadsto       | l = \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \frac {\d y} {\d x} \rd x       | r = C       | c = Linear Combination of Integrals }} {{eqn | ll= \leadsto       | l = \int \frac {\map {g_1} x} {\map {g_2} x} \rd x + \int \frac {\map {h_2} y} {\map {h_1} y} \rd y       | r = C       | c = Integration by Substitution }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k       | r = a \sum_{k \mathop = 0}^{n - 1} r^k + d \sum_{k \mathop = 0}^{n - 1} k r^k       | c =  }} {{eqn | r = \frac {a \paren {1 - r^n} } {1 - r} + d \sum_{k \mathop = 0}^{n - 1} k r^k       | c = Sum of Geometric Sequence }} {{eqn | r = \frac {a \paren {1 - r^n} } {1 - r} + d \paren {\frac {\paren {n - 1} r^{n + 1} - n r^n + r} {\paren {r - 1}^2} }       | c = Sum of Sequence of Power by Index }} {{end-eqn}} Hence the result, after algebra. {{qed}}
From Sum of Arithmetic-Geometric Sequence, we have: :$\displaystyle s_n = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$ We have that $\size r < 1$. So by Sequence of Powers of Number less than One: :$r^n \to 0$ as $n \to \infty$ and :$r^{n - 1} \to 0$ as $n - 1 \to \infty$ Hence: :$s_n \to \dfrac a {1 - r} + \dfrac {r d} {\paren {1 - r}^2}$ as $n \to \infty$. The result follows. {{qed}}
With a view to expressing the problem in the form: :$\displaystyle \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {rd u} {\rd x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = \cos^{n - 1} x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\rd u} {\rd x}       | r = -\paren {n - 1} \cos ^{n - 2} x \sin x       | c = Chain Rule for Derivatives, Derivative of Cosine Function, Derivative of Power }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\rd v} {\rd x}       | r = \cos x       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \sin x       | c = Primitive of Cosine Function }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \cos^n x \rd x       | r = \int \cos^{n - 1} x \cos x \rd x       | c =  }} {{eqn | r = \cos^{n - 1} x \sin x - \int \sin x \paren {-\paren {n - 1} \cos^{n - 2} x \sin x} \rd x       | c = Integration by Parts }} {{eqn | r = \cos^{n - 1} x \sin x + \int \paren {n - 1} \cos^{n - 2} x \sin^2 x \rd x       | c = rearranging }} {{eqn | r = \cos^{n - 1} x \sin x + \int \paren {n - 1} \cos^{n - 2} x \paren {1 - \cos^2 x} \rd x       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \cos^{n - 1} x \sin x + \paren {n - 1} \int \cos^{n - 2} x \rd x - \int \paren {n - 1} \cos^n x \rd x       | c = Linear Combination of Integrals }} {{eqn | ll= \leadsto       | l = n \int \cos^n x \rd x       | r = \cos^{n - 1} x \sin x + \paren {n - 1} \int \cos^{n - 2} x \rd x       | c = rearranging }} {{eqn | ll= \leadsto       | l = \int \cos^n x \rd x       | r = \frac {\cos^{n - 1} x \sin x} n + \frac {n - 1} n \int \cos^{n - 2} x \rd x       | c = dividing both sides by $n$ }} {{end-eqn}} {{qed}}
First: {{begin-eqn}} {{eqn | l = a x^2 + b x + c       | r = \frac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}       | c = Completing the Square }} {{eqn | n = 1       | ll= \leadsto       | l = \int \frac {\d x} {\sqrt {a x^2 + b x + c} }       | r = \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }       | c =  }} {{end-eqn}} Put: {{begin-eqn}} {{eqn | l = z       | r = 2 a x + b       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 a       | c = Derivative of Power }} {{end-eqn}} Let $D = b^2 - 4 a c$. Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {a x^2 + b x + c} }       | r = \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }       | c = from $(1)$ }} {{eqn | r = \int \frac {\d z} {\sqrt a \sqrt {z^2 - D} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - D} }       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} === Negative Discriminant === {{:Primitive of Reciprocal of Root of a x squared plus b x plus c/Negative Discriminant}} {{qed|lemma}} === Positive Discriminant === {{:Primitive of Reciprocal of Root of a x squared plus b x plus c/Positive Discriminant}} {{qed|lemma}} === Zero Discriminant === {{:Primitive of Reciprocal of Root of a x squared plus b x plus c/Zero Discriminant}} {{qed}}
Let $b^2 - 4 a c < 0$. Then: {{begin-eqn}} {{eqn | l = - D       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = - D       | r = q^2       | c = for some $q \in \R$ }} {{eqn | ll= \leadsto       | l = q       | r = \sqrt {4 a c - b^2}       | c = by definition of $D$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt{a x^2 + b x + c} }       | r = \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 + q^2} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 {\sqrt a} \sinh^{-1} {\frac z q} + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ }} {{eqn | r = \frac 1 {\sqrt a} \sinh^{-1} {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C       | c = substituting for $z$ and $q$ }} {{end-eqn}}
Let $b^2 - 4 a c > 0$. Then: {{begin-eqn}} {{eqn | l = D       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = D       | r = q^2       | c = for some $q \in \R$ }} {{eqn | ll= \leadsto       | l = q       | r = \sqrt {b^2 - 4 a c}       | c = by definition of $D$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt{a x^2 + b x + c} }       | r = \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - q^2} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 {\sqrt a} \map \ln {z + \sqrt {z^2 - q^2} } + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ }} {{eqn | r = \frac 1 {\sqrt a} \map \ln {2 a x + b + \sqrt {\paren {2 a x + b}^2 - b^2 - 4 a c} } + C       | c = substituting for $z$ and $q$ }} {{eqn | r = \frac 1 {\sqrt a} \map \ln {2 a x + b + \sqrt {4 a \paren {a x^2 + b x + c} } } + C       | c = Completing the Square }} {{eqn | r = \frac 1 {\sqrt a} \map \ln {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C       | c = rearranging }} {{end-eqn}}
Let $b^2 - 4 a c = 0$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {a x^2 + b x + c} }       | r = \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2} }       | c = from $(1)$ }} {{eqn | r = 2 \sqrt a \int \frac {\d x} {2 a x + b}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = 2 \sqrt a \frac 1 {2 a} \ln \size {2 a x + b} + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | r = \frac 1 {\sqrt a} \ln \size {2 a x + b} + C       | c = simplifying }} {{end-eqn}}
{{begin-eqn}} {{eqn | l = a       | r = 0       | c =  }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | r = \int \frac {\mathrm d x} {\sqrt {b x + c} }       | c =  }} {{eqn | r = \frac {2 \sqrt {b x + c} } b       | c = Primitive of $\dfrac 1 {\sqrt {a x + b}} $ }} {{end-eqn}} {{qed}} Category:Primitive of Reciprocal of Root of a x squared plus b x plus c h11pl3nu2sw5xfhcp0ruub09bcgxjon
First: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {\d x} {a x^2 + c}       | c =  }} {{eqn | r = \frac 1 a \int \frac {\d x} {x^2 + \frac c a}       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} Let $a c > 0$. Then $\dfrac c a > 0$ and: {{begin-eqn}} {{eqn | l = \frac c a       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \frac c a       | r = q^2       | c = for some $q \in \R$ }} {{eqn | ll= \leadsto       | l = q       | r = \sqrt {\frac c a} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \frac 1 a \int \frac {\d x} {x^2 + q^2}       | c = Integration by Substitution }} {{eqn | r = \frac 1 {a q} \arctan \frac x q + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac 1 {a \sqrt {\frac c a} } \arctan \frac x {\sqrt {\frac c a} } + C       | c = substituting for $q$ }} {{eqn | r = \dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C       | c = simplifying }} {{end-eqn}} {{qed|lemma}} Let $a c < 0$. Then $\dfrac c a > 0$ and: {{begin-eqn}} {{eqn | l = \frac c a       | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = -\frac c a       | r = q^2       | c = for some $q \in \R$ }} {{eqn | ll= \leadsto       | l = q       | r = \sqrt {-\frac c a} }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \frac 1 a \int \frac {\d x} {x^2 - q^2}       | c =  }} {{eqn | r = \frac 1 {2 a q} \ln \size {\frac {z - q} {z + q} } + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$ }} {{eqn | r = \frac 1 {2 a \sqrt {-\frac c a} } \ln \size {\frac {z - \sqrt {-\frac c a} } {z + \sqrt {-\frac c a} } } + C       | c = substituting for $q$ }} {{eqn | r = \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C       | c = simplifying }} {{end-eqn}} {{qed|lemma}} Let $c = 0$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a x^2 + b x + c}       | r = \int \frac {\d x} {a x^2}       | c = }} {{eqn | r = \dfrac {-1} {a x} + C       | c = Primitive of Power }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \int \frac {\d x} {x^2 \sqrt {a x^2 + b x + c} } }} {{eqn | r = \int \frac {c \d x} {c x^2 \sqrt {a x^2 + b x + c} }       | c = multiplying top and bottom by $c$ }} {{eqn | r = \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x^2 \sqrt {a x^2 + b x + c} }       | c = adding and subtracting $a x^2 + b x$ }} {{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x^2 \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {x^2 \rd x} {x^2 \sqrt {a x^2 + b x + c} } - \frac b c \int \frac {x \rd x} {x^2 \sqrt {a x^2 + b x + c} }       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 c \int \frac {\sqrt {a x^2 + b x + c} \rd x} {x^2} - \frac a c \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac b c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }       | c = simplification }} {{eqn | r = \frac 1 c \paren {-\frac {\sqrt {a x^2 + b x + c} } x + a \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + \frac b 2 \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } }       | c = Primitive of $\dfrac {\sqrt {a x^2 + b x + c} } {x^2}$ }} {{eqn | o =        | ro= -       | r = \frac a c \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac b c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | r = -\frac {\sqrt {a x^2 + b x + c} } {c x} - \frac b {2 c} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }       | c = gathering terms }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = \paren {2 a x + b}^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 4 a \paren {2 a x + b}       | c = Derivative of Power and Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 4 a \sqrt z       | c = Derivative of Power and Chain Rule for Derivatives }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | o =        | r = \int \sqrt {a x^2 + b x + c} \rd x       | c =  }} {{eqn | r = \int \sqrt {\frac {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } {4 a} } \rd x       | c = Completing the Square }} {{eqn | r = \int \frac {\sqrt {z + \paren {4 a c - b^2} } \rd z} {\paren {2 \sqrt a} \paren {4 a \sqrt z} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 {8 a \sqrt a} \int \frac {\sqrt {z + \paren {4 a c - b^2} } \rd z} {\sqrt z}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 {8 a \sqrt a} \paren {\sqrt z \sqrt {z + \paren {4 a c - b^2} } + \frac {4 a c - b^2} 2 \int \frac {\d z} {\sqrt z \sqrt {z + \paren {4 a c - b^2} } } }       | c = Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$ }} {{eqn | r = \frac {\sqrt z \sqrt {z + \paren {4 a c - b^2} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {\d z} {\sqrt z \sqrt {z + \paren {4 a c - b^2} } }       | c = multiplying out }} {{eqn | r = \frac {\paren {2 a x + b} \sqrt {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {4 a \rd x} {\sqrt {\paren {2 a x + b}^2 + \paren {4 a c - b^2} } }       | c = substituting back for $z$ and $\d z$ }} {{eqn | r = \frac {\paren {2 a x + b} \paren {2 \sqrt a \sqrt {a x^2 + b x + c} } } {8 a \sqrt a} + \frac {4 a c - b^2} {16 a \sqrt a} \int \frac {4 a \rd x} {2 \sqrt a \sqrt {a x^2 + b x + c} }       | c = substituting back for $\sqrt {a x^2 + b x + c}$ }} {{eqn | r = \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \int x \sqrt {a x^2 + b x + c} \ \mathrm d x }} {{eqn | r = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac b {2 a} \int \sqrt {a x^2 + b x + c} \ \mathrm d x       | c = Lemma for Primitive of $x \sqrt {a x^2 + b x + c}$ }} {{eqn | r = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a}       | c = }} {{eqn | o =        | ro= -       | r = \frac b {2 a} \left({\frac {\left({2 a x + b}\right) \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} } }\right)       | c = Primitive of $\sqrt {a x^2 + b x + c}$ }} {{eqn | r = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a}       | c = }} {{eqn | o =        | ro= -       | r = \frac {b \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \left({4 a c - b^2}\right)} {16 a^2} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \int \frac {\sqrt {a x^2 + b x + c} } x \rd x       | o =  }} {{eqn | r = \int \frac {a x^2 + b x + c} {x \sqrt {a x^2 + b x + c} } \rd x       | c = multiplying top and bottom by $\sqrt {a x^2 + b x + c}$ }} {{eqn | r = a \int \frac {x^2 \rd x} {x \sqrt {a x^2 + b x + c} } + b \int \frac {x \rd x} {x \sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }       | c = Linear Combination of Integrals }} {{eqn | r = a \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} } + b \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }       | c = simplifying }} {{eqn | r = a \paren {\frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } }       | c = Primitive of $\dfrac x {\sqrt {a x^2 + b x + c} }$ }} {{eqn | o =        | ro= +       | r = b \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | r = \sqrt {a x^2 + b x + c} + \frac b 2 \int \frac {\d x} {\sqrt {a x^2 + b x + c} } + c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \int \frac {\sqrt {a x^2 + b x + c} } {x^2} \ \mathrm d x       | o =  }} {{eqn | r = \int \frac {a x^2 + b x + c} {x^2 \sqrt {a x^2 + b x + c} } \ \mathrm d x       | c = multiplying top and bottom by $\sqrt {a x^2 + b x + c}$ }} {{eqn | r = a \int \frac {x^2 \ \mathrm d x} {x^2 \sqrt {a x^2 + b x + c} } + b \int \frac {x \ \mathrm d x} {x^2 \sqrt {a x^2 + b x + c} } + c \int \frac {\mathrm d x} {x^2 \sqrt {a x^2 + b x + c} }       | c = Linear Combination of Integrals }} {{eqn | r = a \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} } + b \int \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} } + c \int \frac {\mathrm d x} {x^2 \sqrt {a x^2 + b x + c} }       | c = simplifying }} {{eqn | r = c \left({-\frac {\sqrt {a x^2 + b x + c} } {c x} - \frac b {2 c} \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} } }\right)       | c = Primitive of $\dfrac 1 {x^2 \sqrt {a x^2 + b x + c} }$ }} {{eqn | o =        | ro= +       | r = a \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} } + b \int \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | r = \frac {-\sqrt {a x^2 + b x + c} } x + a \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} } + \frac b 2 \int \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} }       | c = simplifying }} {{end-eqn}} {{qed}}