hcju/proofwikiqa huggingface dataset:
Let $A, B \in \powerset S$. Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$. We also have $A \setminus B \subseteq A$ from Set Difference is Subset. Thus by Subset Relation is Transitive, $A \setminus B \subseteq S$. Thus $A \setminus B \in \powerset S$, and closure is proved. {{qed}}
From Index Laws for Semigroup: Sum of Indices: :$\forall m, n \in \Z_{>0}: a^{n + m} = a^n \odot a^m$ But from Integer Addition is Commutative: :$n + m = m + n$ So: :$a^n \odot a^m = a^{n + m} = a^{m + n} = a^m \odot a^n$ {{Qed}}
Note that the Power Rule cannot be used because the index is not a constant. {{begin-eqn}} {{eqn | l = \frac \d {\d x} x^x       | r = \frac \d {\d x} \map \exp {x \ln x}       | c = {{Defof|Power (Algebra)|subdef = Real Number|index = 1}} }} {{eqn | r = \paren {\frac \d {\map \d {x \ln x} } \map \exp {x \ln x} } \paren {\map {\frac \d {\d x} } {x \ln x} }       | c = Chain Rule for Derivatives }} {{eqn | r = \map \exp {x \ln x} \paren {\map {\frac \d {\d x} } {x \ln x} }       | c = Derivative of Exponential Function }} {{eqn | r = \map \exp {x \ln x} \paren {\map {\frac \d {\d x} } x \cdot \ln x + x \frac \d {\d x} \ln x}       | c = Product Rule for Derivatives }} {{eqn | r = \map \exp {x \ln x} \paren {1 \cdot \ln x + x \cdot \frac 1 x}       | c = Derivative of Identity Function and Derivative of Natural Logarithm Function }} {{eqn | r = \map \exp {x \ln x} \paren {\ln x + 1}       | c = Real Multiplication Identity is One and Inverses for Real Multiplication }} {{eqn | r = x^x \paren {\ln x + 1}       | c = {{Defof|Power (Algebra)|subdef = Real Number|index = 1}} }} {{end-eqn}} {{qed}}
Note that the Power Rule cannot be used because the index is not a constant. Let $y := x^x$. As $x$ was stipulated to be positive, we can take the natural logarithm of both sides: {{begin-eqn}} {{eqn | l = \ln y       | r = \ln x^x }} {{eqn | r = x \ln x       | c = Laws of Logarithms }} {{eqn | l = \map {\frac \d {\d x} } {\ln y}       | r = \map {\frac \d {\d x} } {x \ln x}       | c = differentiating both sides {{WRT|Differentiation}} $x$ }} {{eqn | l = \frac 1 y \frac {\d y} {\d x}       | r = \map {\frac \d {\d x} } {x \ln x}       | c = Chain Rule for Derivatives and Derivative of Natural Logarithm Function }} {{eqn | r = \map {\frac \d {\d x} } x \cdot \ln x + x \frac {\d} {\d x} \ln x       | c = Product Rule for Derivatives }} {{eqn | r = 1 \cdot \ln x + x \cdot \frac 1 x       | c = Derivative of Identity Function and Derivative of Natural Logarithm Function }} {{eqn | r = \ln x + 1       | c = simplification }} {{eqn | l = \frac {\d y} {\d x}       | r = x^x \paren {\ln x + 1}       | c = multiplying both sides by $y = x^x$ }} {{end-eqn}} {{qed}}
From Derivative of $x^{a x}$ we have: :$\dfrac \d {\d x} x^{a x} = a x^{a x} \paren {\ln x + 1}$ The result follows on setting $a = 1$. {{qed}}
Let $a, b \in S: a \circ b = b \circ a$. Then from Powers of Commuting Elements of Semigroup Commute: :$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$ {{qed|lemma}} However, consider the dihedral group $D_3$ $= \gen {a, b: a^3 = b^2 = e, b a b = a^{−1} }$. A group is a semigroup. Moreover, the Cancellation Laws hold in a group. Hence both $a$ and $b$ are cancellable. We have $a^3 \circ b^2 = b^2 \circ a^3 = e \circ e = e$. However $a \circ b \ne b \circ a$. Hence the result by Proof by Counterexample. {{qed}} Category:Semigroups Category:Commutativity 5mm0or7how3n76ul1oqe1k6fzc5bf1l
=== Necessary Condition === Let $x \circ y = y \circ x$. Then by Power of Product of Commuting Elements in Semigroup equals Product of Powers: :$\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$ {{qed|lemma}} === Sufficient Condition === Suppose $\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$. In particular, when $n = 2$, {{begin-eqn}} {{eqn | l = \paren {x \circ y}^2       | r = x^n \circ y^2       | c =  }} {{eqn | l = x \circ y \circ x \circ y       | r = x \circ x \circ y \circ y       | c = {{Defof|Semigroup}}: $\circ$ is Associative }} {{eqn | l = y \circ x       | r = x \circ y       | c = $x$ and $y$ are assumed to be cancellable }} {{end-eqn}} {{qed}} {{explain|This needs to go into an inductive argument, and the assumption about cancellability needs to be found and utilised explicitly}}
Let $y \in \R$ such that: : $\exists x \in \R, \left|{x}\right| \ge 1: x = \csc \left({y + \dfrac \pi 2}\right)$ Then: {{begin-eqn}} {{eqn | l = x       | r = \sec \left({y + \frac \pi 2}\right)       | c =  }} {{eqn | r = -\csc y       | c = Secant of Angle plus Right Angle }} {{eqn | r = \csc \left({-y}\right)       | c = Cosecant Function is Odd }} {{end-eqn}} Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$. Then we can write $-y = \operatorname{arccsc} x$. But then $\csc \left({y + \dfrac \pi 2}\right) = x$. Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$. Hence $y + \dfrac \pi 2 = \operatorname{arcsec} x$. That is, $\dfrac \pi 2 = \operatorname{arcsec} x + \operatorname{arccosec} x$. {{qed}}
=== Non-Vertical Tangent Lines === From Equation of Circle, $\mathcal C$ can be described on the $x y$-plane in the form: :$\paren {x - a}^2 + \paren {y - b}^2 = r^2$ where $P = \tuple {a, b}$ is the center of the circle and $r$ is the radius.  We use the definition of the derivative as the gradient of the tangent line $\mathcal T$. Taking the derivative {{WRT|Differentiation}} $x$ of both sides of the equation we get: {{begin-eqn}} {{eqn | l = 2 \paren {x - a} + 2 \paren {y - b} \frac {\d y} {\d x}       | r = 0       | c = Derivative of Constant, Chain Rule for Derivatives, Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {a - x} {y - b}       | c = }} {{end-eqn}} This is the slope at any point on $\mathcal C$. From the slope-intercept form of a line, the equation of such a line is: :$y - y_n = m \paren {x - x_n}$ given any point $\paren {x_n, y_n}$ and the gradient $m$. For $\mathcal T$: :$m = \left.{\dfrac {\d y} {\d x} }\right\vert^{x = x_n}_{y = y_n}$ Thus the equation of $\mathcal T$ is: :$y - y_n = \dfrac {a - x_n} {y_n - b} \paren {x - x_n}$ {{qed|lemma}} === Vertical Tangent Lines === {{ProofWanted|by the Infinite Limit Theorem}}
We have $a^0 = e$ so it follows trivially that $a^{-0} = \paren {a^{-1} }^0$. From the general inverse of product, we have: :$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ where $a_1, a_2, \ldots, a_n \in S$ are all invertible for $\circ$. So we can put $a_1, a_2, \ldots, a_n = a$ and we have that : $a^n$ is invertible for all $n \in \N$ : $\forall n \in \N: \paren {a^n}^{-1} = \paren {a^{-1 }}^n$ From the above: :$a^{-n} = \paren {a^{-1} }^n$ Thus: {{begin-eqn}} {{eqn | l = \paren {a^{-n} }^{-1}       | r = \paren {\paren {a^{-1} }^n}^{-1}       | c =  }} {{eqn | r = \paren {\paren {a^{-1} }^{-1} }^n       | c =  }} {{eqn | r = a^{-\paren {-n} }       | c =  }} {{end-eqn}} Similarly, if $a$ is invertible then $a^{-1}$ is also invertible. So we also have: :$\circ^{-n} \paren {a^{-1} } = \circ^n \paren {\paren {a^{-1} }^{-1} }$ Thus: {{begin-eqn}} {{eqn | l = a^{-\paren {-n} }       | r = \paren {\paren {a^{-1} }^{-1} }^n       | c =  }} {{eqn | r = \paren {a^{-1} }^{-n}       | c =  }} {{end-eqn}} Thus the result holds for all $n \in \Z$. {{qed}}
To prove the claim, it will suffice to show that for each $N \in \R_{>0}$, one can find an $\epsilon > 0$ such that: :$\size {x - c} < \epsilon \implies \size {\map f x} \ge N$ So fix $N \in \R_{>0}$. First, by continuity of $g$, find an $\epsilon_1 > 0$ such that: :$\size {x - c} < \epsilon_1 \implies \size {\map g x - \map g c} < \size {\dfrac {\map g c} 2}$ which by the corollary to the Reverse Triangle Inequality implies that: :$\size {x - c} < \epsilon_1 \implies \size {\map g c} - \size {\map g x} < \size {\dfrac {\map g c} 2}$ and by rearranging, this becomes: :$\size {x - c} < \epsilon_1 \implies \size {\map g x} > \size {\dfrac {\map g c} 2}$ Next, by continuity of $h$, find an $\epsilon_2 > 0$ such that $\epsilon_1 > \epsilon_2$ and: :$\size {x - c} < \epsilon_2 \implies \size {\map h x} < \dfrac {\size {\map g c} } {2 N}$ From Ordering of Reciprocals, this means: :$\size {x - c} < \epsilon_2 \implies \dfrac 1 {\size {\map h x} } > \dfrac {2 N} {\size {\map g c} }$ These together imply that, whenever $0 < \size {x - c} < \epsilon_2$, we have: {{begin-eqn}} {{eqn | l = \size {\map f x}       | r = \size {\frac {\map g x} {\map h x)} } }} {{eqn | r = \size {\map g x} \cdot \size {\frac 1 {\map h x} } }} {{eqn | o = >       | r = \size {\frac {\map g c} 2} \cdot \frac 1 {\size {\map h x} }       | c = since $\epsilon_2 < \epsilon_1$ }} {{eqn | o = >       | r = \size {\frac {\map g c} 2} \cdot \frac {2 N} {\size {\map g c} } }} {{eqn | r = N }} {{end-eqn}} Thus $\epsilon_2$ is a valid choice for the sought $\epsilon$. Hence the result. {{qed}}
Let $m \in \N, c = a^m, d = \paren {a^{-1}}^m$. We define the mapping $g_c: \Z \to S$ as: :$\forall n \in \Z: \map {g_c} n = \map {\circ^n} c$ as defined in the proof of the Index Law for Sum of Indices. Let $h: \Z \to \Z$ be the mapping defined as: :$\forall z \in \Z: \map h z = z m$ Then: {{begin-eqn}} {{eqn | l = a^{n m}       | r = \map {\paren {g_a \circ h} } n }} {{eqn | l = \paren {a^m}^n       | r = \map {g_c} n }} {{end-eqn}} By Index Law for Sum of Indices and Index Laws for Semigroup: Product of Indices, $g_a \circ h$ and $g_c$ are homomorphisms from $\Z$ to $S$ which coincide on $\N$. So by the Extension Theorem for Homomorphisms, $g_a \circ h = g_c$. Therefore: :$\forall n \in \Z, m \in \N: a^{n m} = \paren {a^m}^n$ Also: {{begin-eqn}} {{eqn | l = a^{n \paren {-m} }       | r = a^{-\paren {n m} } }} {{eqn | r = \paren {a^{-1} }^{n m} }} {{eqn | r = \map {\paren {g_{a^{-1} } \circ h} } n }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = \paren {a^{-m} }^n       | r = \paren {\paren {a^{-1} }^m}^n }} {{eqn | r = \map {g_d} n }} {{end-eqn}} So, by the same reasoning as before, $g_{a^{-1}} \circ h = g_d$. Therefore: :$\forall n \in \Z, m \in \N: a^{n \paren {-m}} = \paren {a^{-m}}^n$ Thus: :$\forall n, m \in \Z: a^{n m} = \paren {a^m}^n$ As $n m = m n$, the result follows. {{qed}}
For each $c \in S$ which is invertible for $\circ$, we define the mapping $g_c: \Z \to S$ as: :$\forall n \in \Z: \map {g_c} n = \map {\circ^n} c$ By the Index Law for Monoids: Negative Index, $\map {g_a} n$ is invertible for all $n \in \Z$. By definition of Power of Element of Monoid, the restriction of $g_a$ to $\N$ is a homomorphism from $\struct {\N, +}$ to $\struct {S, \circ}$. From the definition of Power of Element of Monoid, $\map {g_a} 0$ is the identity for $\circ$. Hence, by the Extension Theorem for Homomorphisms, there is a unique homomorphism $h_a: \paren {\N, +} \to \paren {S, \circ}$ which coincides in $\N$ with $g_c$. But by Homomorphism with Identity Preserves Inverses: {{begin-eqn}} {{eqn | lo= \forall n > 0:       | l = \map {h_a} {-n}       | r = \paren {\map {h_a} n}^{-1}  }} {{eqn | r = \paren {a^n}^{-1}       | c =  }} {{eqn | r = a^{-n}       | c =  }} {{eqn | r = \map {g_a} {-n}       | c =  }} {{end-eqn}} Hence $h_a = g_a$ and so $g_a$ is a homomorphism and so the result follows. {{qed}}
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices: :$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$ {{qed}}
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices: :$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$ {{qed}}
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Product of Indices: :$\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$ {{qed}}
By definition of inverse element: :$x \circ x^{-1} = e = x^{-1} \circ x$ Hence the result by definition.
By Powers of Commuting Elements of Semigroup Commute, if $m > 0$ and $n > 0$ then $a^m$ commutes with $b^n$. By Commutation with Inverse in Monoid, again if $m > 0$ and $n > 0$ then $a^m$ commutes with $\paren {b^n}^{-1} = b^{-n}$. Similarly $b^n$ commutes with $a^{-m}$. But as $a^{-m}$ commutes with $b^n$, it also commutes with $\paren {b^n}^{-1} = b^{-n}$, again by Commutation with Inverse in Monoid. Hence the result. {{qed}}
From Power of Product of Commutative Elements in Semigroup, this result holds if $n \ge 0$. Since $a$ and $b$ commute, then so do $a^{-1}$ and $b^{-1}$ by Commutation of Inverses in Monoid. Hence, if $n > 0$: {{begin-eqn}} {{eqn | l = \paren {a \circ b}^{-n}       | r = \paren {\paren {a \circ b}^{-1} }^n }} {{eqn | r = \paren {b^{-1} \circ a^{-1} }^n       | c = Inverse of Product }} {{eqn | r = \paren {a^{-1} \circ b^{-1} }^n       | c = Commutation of Inverses in Monoid }} {{eqn | r = \paren {a^{-1} }^n \circ \paren {b^{-1} }^n       | c =  }} {{eqn | r = a^{-n} \circ b^{-n}       | c =  }} {{end-eqn}} The result follows. {{qed}}
By definition, all elements of a group are invertible. Therefore Commutativity of Powers in Monoid can be applied directly. {{Qed}} Category:Group Theory Category:Commutativity s8ogf503onvplgoz7ir9lu0b4wljf2f
=== Necessary Condition === Let $a \circ b = b \circ a$. By definition, all elements of a group are invertible. Therefore the results in Power of Product of Commutative Elements in Monoid can be applied directly. {{qed|lemma}} === Sufficient Condition === If $\paren {a \circ b}^n = a^n \circ b^n$ for all $n$, then it certainly holds for $n = 2$: {{begin-eqn}} {{eqn | l = \paren {a \circ b}^2       | r = a^2 \circ b^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a \circ b} \circ \paren {a \circ b}       | r = \paren {a \circ a} \circ \paren {b \circ b}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a^{-1} \circ a} \circ b \circ a \circ \paren {b \circ b^{-1} }       | r = \paren {a^{-1} \circ a} \circ a \circ b \circ \paren {b \circ b^{-1} }       | c = {{GroupAxiom|1}} }} {{eqn | ll= \leadsto       | l = e \circ b \circ a \circ e       | r = e \circ a \circ b \circ e       | c = {{GroupAxiom|3}} }} {{eqn | ll= \leadsto       | l = b \circ a       | r = a \circ b       | c = {{GroupAxiom|2}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \tan x \rd x       | r = \int \frac {\sin x} {\cos x} \rd x       | c = {{Defof|Real Tangent Function}} }} {{eqn | r = -\int \frac {-\sin x} {\cos x} \rd x       | c = Multiply by $1 = \dfrac {-1} {-1}$ }} {{eqn | r = -\int \frac {\paren {\cos x}'} {\cos x} \rd x       | c = Derivative of Cosine Function }} {{eqn | r = -\ln \size {\cos x} + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
Suppose $x > 0$. Then: :$\ln \size x = \ln x$ The result follows from Derivative of Natural Logarithm Function and the definition of primitive. Suppose $x < 0$. Then: {{begin-eqn}} {{eqn | l = \dfrac \d {\d x} \ln \size x       | r = \dfrac \d {\d x} \map \ln {-x}       | c = {{Defof|Absolute Value}} }} {{eqn | r = \frac 1 {-x} \cdot -1       | c = Chain Rule for Derivatives and Derivative of Natural Logarithm Function, as $-x > 0$ }} {{eqn | r = \frac 1 x }} {{end-eqn}} and the result again follows from the definition of the primitive. {{qed}}
Let $u = a x + b$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^2}       | r = \frac 1 a \int \frac {\mathrm d u} {u^2}       | c = Primitive of Function of $a x + b$ }} {{eqn | r = \frac 1 a \frac {-1} u + C       | c = Primitive of Power }} {{eqn | r = -\frac 1 {a \left({a x + b}\right)} + C       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
From Primitive of Power of $a x + b$: :$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$ where $n \ne 1$. The result follows by setting $n = -2$. {{qed}}
Let $u = a x + b$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^3}       | r = \frac 1 a \int \frac {\mathrm d u} {u^3}       | c = Primitive of Function of $a x + b$ }} {{eqn | r = \frac 1 a \frac {-1} {2 u^2} + C       | c = Primitive of Power }} {{eqn | r = -\frac 1 {2 a \left({a x + b}\right)^2} + C       | c = substituting for $u$ }} {{end-eqn}} {{qed}}
From Primitive of Power of $a x + b$: :$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$ where $n \ne 1$. The result follows by setting $n = -3$. {{qed}}
From Primitive of Reciprocal of Power of x by Power of a x squared plus b x plus c: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^m \paren {a x^2 + b x + c}^n}       | r = \frac {-1} {\paren {m - 1} c x^{m - 1} \paren {a x^2 + b x + c}^{n - 1} } }} {{eqn | o =       | ro= -       | r = \frac {\paren {m + 2 n - 3} a} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 2} \paren {a x^2 + b x + c}^n} }} {{eqn | o =       | ro= -       | r = \frac {\paren {m - n + 2} b} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 1} \paren {a x^2 + b x + c}^n} }} {{end-eqn}} Setting $m = n = 2$: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}       | r = \frac {-1} {\paren {2 - 1} c x^{2 - 1} \paren {a x^2 + b x + c}^{2 - 1} }       | c =  }} {{eqn | o =        | ro= -       | r = \frac {\paren {2 + 2 \times 2 - 3} a} {\paren {2 - 1} c} \int \frac {\d x} {x^{2 - 2} \paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | o =        | ro= -       | r = \frac {\paren {2 - 2 + 2} b} {\paren {2 - 1} c} \int \frac {\d x} {x^{2 - 1} \paren {a x^2 + b x + c}^2}       | c =  }} {{eqn | r = \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \paren {a x + b} }       | r = \int \paren {\dfrac 1 {b x} - \dfrac a {b \paren {a x + b} } } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 b \int \frac {\d x} x - \frac a b \int \frac {\d x} {a x + b}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 b \ln \size x - \frac a b \int \frac {\d x} {a x + b} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 b \ln \size x - \frac 1 b \ln \size {a x + b} + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \paren {a x + b} }       | r = \int \frac {b \rd x} {b x \paren {a x + b} }       | c = multiplying top and bottom by $b$ }} {{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b} }       | c = adding and subtracting $a x$ }} {{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x \paren {a x + b} }       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 b \int \frac {\d x} x - \frac a b \int \frac {\d x} {a x + b}       | c = simplifying }} {{eqn | r = \frac 1 b \ln \size x - \frac a b \int \frac {\d x} {a x + b} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 b \ln \size x - \frac 1 b \ln \size {a x + b} + C       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {a x + b} }       | r = \int \paren {-\frac a {b^2 x} + \frac 1 {b x^2} + \frac {a^2} {b^2 \paren {a x + b} } } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = -\frac a {b^2} \int \frac {\d x} x + \frac 1 b \int \frac {\d x} {x^2} + \frac {a^2} {b^2} \int \frac {\d x} {a x + b}       | c = Linear Combination of Integrals }} {{eqn | r = -\frac a {b^2} \int \frac {\d x} x + \frac 1 b \frac {-1} x + \frac {a^2} {b^2} \int \frac {\d x} {a x + b} + C       | c = Primitive of Power }} {{eqn | r = -\frac a {b^2} \ln \size x - \frac 1 {b x} + \frac {a^2} {b^2} \int \frac {\d x} {a x + b} + C       | c = Primitive of Reciprocal }} {{eqn | r = -\frac a {b^2} \ln \size x - \frac 1 {b x} + \frac a {b^2} \ln \size {a x + b} + C       | c = Primitive of Reciprocal of a x + b }} {{eqn | r = -\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {a x + b} }       | r = \int \frac {b \rd x} {b x^2 \paren {a x + b} }       | c = multiplying top and bottom by $b$ }} {{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x^2 \paren {a x + b} }       | c = adding and subtracting $a x$ }} {{eqn | r = \frac 1 b \int \frac {\paren {a x + b } \rd x} {x^2 \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x^2 \paren {a x + b} }       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 b \int \frac {\d x} {x^2} - \frac a b \int \frac {\d x} {x \paren {a x + b} }       | c = simplification }} {{eqn | r = -\frac 1 {b x} - \frac a b \int \frac {\d x} {x \paren {a x + b} } + C       | c = Primitive of Power }} {{eqn | r = -\frac 1 {b x} - \frac a b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } + C       | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$ }} {{eqn | r = -\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} + C       | c = Logarithm of Reciprocal and simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^3 \left({a x + b}\right)}       | r = \int \frac {b \ \mathrm d x} {b x^3 \left({a x + b}\right)}       | c = multiplying top and bottom by $b$ }} {{eqn | r = \int \frac {\left({a x + b - a x}\right) \ \mathrm d x} {b x^3 \left({a x + b}\right)}       | c = adding and subtracting $a x$ }} {{eqn | r = \frac 1 b \int \frac {\left({a x + b}\right) \ \mathrm d x} {x^3 \left({a x + b}\right)} - \frac a b \int \frac {x \ \mathrm d x} {x^3 \left({a x + b}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 b \int \frac {\mathrm d x} {x^3} - \frac a b \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)}       | c = simplification }} {{eqn | r = -\frac 1 {2 b x^2} - \frac a b \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)} + C       | c = Primitive of Power }} {{eqn | r = -\frac 1 {2 b x^2} - \frac a b \left({-\frac 1 {b x} + \frac a {b^2} \ln \left\vert{\frac {a x + b} x}\right\vert}\right) + C       | c = Primitive of $\dfrac 1 {x^2 \left({a x + b}\right)}$ }} {{eqn | r = -\frac 1 {2 b x^2} + \frac a {b^2 x} - \frac {a^2} {b^3} \ln \left\vert{\frac {a x + b} x}\right\vert + C       | c = multiplying out }} {{eqn | r = \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {a^2} {b^3} \ln \left\vert{\frac x {a x + b} }\right\vert + C       | c = Difference of Logarithms }} {{eqn | r = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \left\vert{\frac x {a x + b} }\right\vert + C       | c = rearranging }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b} }       | r = \int \paren {\frac {a^2} {b^3 x} + \frac {-a} {b^2 x^2} + \frac 1 {b x^3} + \frac {-a^3} {b^3 \paren {a x + b} } } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac {a^2} {b^3} \int \frac {\d x} x + \frac {-a} {b^2} \int \frac {\d x} {x^2} + \frac 1 b \int \frac {\d x} {x^3} + \frac {-a^3} {b^3} \int \frac {\d x} {a x + b}       | c = Linear Combination of Integrals }} {{eqn | r = \frac {a^2} {b^3} \int \frac {\d x} x + \frac {-a} {b^2} \frac {-1} x + \frac 1 b \frac {-1} {2 x^2} + \frac {-a^3} {b^3} \int \frac {\d x} {a x + b} + C       | c = Primitive of Power }} {{eqn | r = \frac {a^2} {b^3} \ln \size x + \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {-a^3} {b^3} \int \frac {\d x} {a x + b} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {a^2} {b^3} \ln \size x + \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {-a^3} {b^3} \frac 1 a \ln \size {a x + b} + C       | c = Primitive of Reciprocal of a x + b }} {{eqn | r = \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C       | c = Difference of Logarithms }} {{eqn | r = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C       | c = rearranging }} {{end-eqn}} {{qed}}