hcju/proofwikiqa huggingface dataset:
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 + a^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c = Derivative of Power }} {{eqn | ll= \leadsto       | l = \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2}       | r = \int \frac {\paren {z - a^2} } {z^2} \frac {\d z} 2       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\d z} z - \frac {a^2} 2 \int \frac {\d z} {z^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 2 \int \frac {\d z} z + \frac {a^2} {2 z} + C       | c = Primitive of Power }} {{eqn | r = \frac 1 2 \ln z + \frac {a^2} {2 z} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^3 \ \mathrm d x} {\left({x^2 + a^2}\right)^2}       | r = \int \frac {x \left({x^2 + a^2 - a^2}\right)} {\left({x^2 + a^2}\right)^2} \ \mathrm d x       | c =  }} {{eqn | r = \int \frac {x \left({x^2 + a^2}\right)} {\left({x^2 + a^2}\right)^2} \ \mathrm d x - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {x \ \mathrm d x} {x^2 + a^2} - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}       | c = simplification }} {{eqn | r = \frac 1 2 \ln \left({x^2 + a^2}\right) - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2} + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | r = \frac 1 2 \ln \left({x^2 + a^2}\right) - a^2 \left({\frac {-1} {2 \left({x^2 - a^2}\right)} }\right) + C       | c = Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$ }} {{eqn | r = \frac {a^2} {2 \left({x^2 + a^2}\right)} + \frac 1 2 \ln \left({x^2 + a^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {x^2 + a^2} }       | r = \int \paren {\frac 1 {a^2 x^2} - \frac 1 {a^2 \paren {x^2 + a^2} } } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2} - \frac 1 {a^2} \int \frac {1 \rd x} {x^2 + a^2}       | c = Linear Combination of Integrals }} {{eqn | r = -\frac 1 {a^2 x} - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2} + C       | c = Primitive of Power }} {{eqn | r = -\frac 1 {a^2 x} - \frac 1 {a^2} \paren {\frac 1 a \arctan \frac x a} + C       | c = Primitive of Reciprocal of $x^2 + a^2$ }} {{eqn | r = -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^3 \left({x^2 + a^2}\right)}       | r = \int \left({\frac 1 {a^2 x^3} - \frac 1 {a^4 x} + \frac x {a^4 \left({x^2 + a^2}\right)} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^3} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 + a^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac {-1} {a^2 x^2} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 + a^2} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 + a^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {-1} {a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \left({\frac 1 2 \ln \left ({x^2 + a^2}\right)}\right) + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | r = -\frac 1 {a^2 x^2} - \frac 1 {2 a^4} \ln \left({x^2}\right) + \frac 1 {2 a^4} \ln \left ({x^2 + a^2}\right) + C       | c = Logarithm of Power }} {{eqn | r = -\frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C       | c = Difference of Logarithms }} {{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 {\d x} {\paren {x^2 + a^2}^2}       | r = \int \frac {a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta       | c = Integration by Substitution }} {{eqn | r = \int \frac {a \sec^2 \theta} {\paren {a^2 \sec^2 \theta}^2} \rd \theta       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac 1 {a^3} \int \frac {\d \theta} {\sec^2 \theta}       | c = simplification }} {{eqn | r = \frac 1 {a^3} \int \cos^2 \theta \rd \theta       | c = {{Defof|Real Secant Function}} }} {{eqn | r = \frac 1 {a^3} \paren {\frac \theta 2 + \frac {\sin 2 \theta} 4} + C       | c = Primitive of Square of Cosine Function }} {{eqn | r = \frac \theta {2 a^3} + \frac 1 {2 a^3} \frac {\tan \theta} {1 + \tan^2 \theta} + C       | c = Tangent Half-Angle Substitution for Sine }} {{eqn | r = \frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a + C       | c = substituting for $\theta$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 + a^2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 x       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}       | r = \int \frac {\mathrm d z} {2 z^2}       | c = Integration by Substitution }} {{eqn | r = -\frac 1 {2 z} + C       | c = Primitive of Power }} {{eqn | r = -\frac 1 {2 \left({x^2 + a^2}\right)} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2}^2}       | r = \int \paren {\frac 1 {a^4 x} - \frac x {a^4 \paren {x^2 + a^2} } - \frac x {a^2 \paren {x^2 + a^2}^2} } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^4} \int \frac {\d x} x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \paren {-\frac 1 {2 \paren {x^2 + a^2} } } + C       | c = Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$ }} {{eqn | r = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C       | c = simplification }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2}       | r = \int \paren {\frac 1 {a^4 x^2} - \frac 1 {a^4 \paren {x^2 + a^2} } - \frac 1 {a^2 \paren {x^2 + a^2}^2} } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \frac {-1} x - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2} + C       | c = Primitive of Power }} {{eqn | r = -\frac 1 {a^4 x} - \frac 1 {a^4} \paren {\frac 1 a \arctan \frac x a} - \frac 1 {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^4 x} - \frac 1 {a^5} \arctan \frac x a - \frac 1 {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 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C       | c = simplification }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^3 \paren {x^2 + a^2}^2}       | r = \int \paren {-\frac 2 {a^6 x} + \frac 1 {a^4 x^3} + \frac {2 x} {a^6 \paren {x^2 + a^2} } + \frac x {a^4 \paren {x^2 + a^2}^2} } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = -\frac 2 {a^6} \int \frac {\d x} x + \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {x \rd x} {x^2 + a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = -\frac 2 {a^6} \ln x + \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {x \rd x} {x^2 + a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = -\frac 2 {a^6} \ln x - \frac 1 {2 a^4 x^2} + \frac 2 {a^6} \int \frac {x \rd x} {x^2 + a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C       | c = Primitive of Power }} {{eqn | r = -\frac 2 {a^6} \ln x - \frac 1 {2 a^4 x^2} + \frac 2 {a^6} \paren {\frac 1 2 \map \ln {x^2 + a^2} } + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{eqn | r = -\frac 2 {a^6} \ln x - \frac 1 {2 a^4 x^2} + \frac 1 {a^6} \map \ln {x^2 + a^2} + \frac 1 {a^4} \paren {-\frac 1 {2 \paren {x^2 + a^2} } } + C       | c = Primitive of $\dfrac x {\paren {x^2 + a^2}^2}$ }} {{eqn | r = -\frac 1 {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 + a^2} } - \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 + a^2} } + C       | c = simplification }} {{end-eqn}} {{qed}}
Aiming for an expression in the form: : $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$ in order to use the technique of Integration by Parts, let: {{begin-eqn}} {{eqn | l = v       | r = x       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d v} {\mathrm d x}       | r = 1       | c = Power Rule for Derivatives }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = u       | r = \frac 1 {\left({x^2 + a^2}\right)^{n - 1} }       | c =  }} {{eqn | r = \left({x^2 + a^2}\right)^{- \left({n - 1}\right)}       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = - 2 \left({n - 1}\right) x \left({x^2 + a^2}\right)^{- \left({n - 1}\right) - 1}       | c = Power Rule for Derivatives and Chain Rule for Derivatives }} {{eqn | r = - 2 \left({n - 1}\right) x \left({x^2 + a^2}\right)^{-n}       | c =  }} {{eqn | r = \frac {- \left({2 n - 2}\right) x} {\left({x^2 + a^2}\right)^n}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }       | r = \frac 1 {\left({x^2 + a^2}\right)^{n - 1} } x - \int x \frac {- \left({2 n - 2}\right) x} {\left({x^2 + a^2}\right)^n} \ \mathrm d x       | c = Integration by Parts }} {{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2} {\left({x^2 + a^2}\right)^n} \ \mathrm d x       | c = simplifying }} {{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 + a^2 - a^2} {\left({x^2 + a^2}\right)^n} \ \mathrm d x       | c =  }} {{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 + a^2} {\left({x^2 + a^2}\right)^n} \ \mathrm d x - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n}       | c = Linear Combination of Integrals }} {{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} } - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n}        | c =  }} {{eqn | ll= \implies       | l = \left({1 - \left({2 n - 2}\right)}\right) \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }       | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n}       | c =  }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n}       | r = \frac x {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 + a^2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \implies       | l = \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^n}       | r = \int \frac {\mathrm d z} {2 z^n}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int z^{-n} \ \mathrm d z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \frac {z^{-n + 1} } {-n + 1} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {2 \left({n - 1}\right) z^{n - 1} } + C       | c = simplifying }} {{eqn | r = \frac {-1} {2 \left({n - 1}\right) \left({x^2 + a^2}\right)^{n - 1} } + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 + a^2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^n}       | r = \int \frac {\mathrm d z} {2 x^2 z^n}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z - a^2}\right) z^n}       | c = Primitive of Constant Multiple of Function and substituting for $x^2$ }} {{end-eqn}} From Reduction Formula for Primitive of $x^m \left({a x + b}\right)^n$: Increment of Power of $x$: :$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \ \mathrm d x$ Let: {{begin-eqn}} {{eqn | l = a       | o = :=       | r = 1       | c =  }} {{eqn | l = b       | o = :=       | r = -a^2       | c =  }} {{eqn | l = x       | o = :=       | r = z       | c =  }} {{eqn | l = m       | o = :=       | r = -n       | c =  }} {{eqn | l = n       | o = :=       | r = -1       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^n}       | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z - a^2}\right) z^n}       | c =  }} {{eqn | r = \frac 1 2 \left({\frac {z^{-n+1} \left({z - a^2}\right)^{-1 + 1} } {-\left({-n + 1}\right) a^2} - \frac {\left({-n + \left({-1}\right) + 2}\right)} {-\left({-n + 1}\right) a^2} \int z^{-n + 1} \left({z - a^2}\right)^{-1} \ \mathrm d z}\right)       | c =  }} {{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 z^{n-1} } + \frac 1 {2 a^2} \int \frac {\mathrm d z} {z^{n - 1} \left({z - a^2}\right)}       | c = simplification }} {{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n-1} } + \frac 1 {2 a^2} \int \frac {2 x \ \mathrm d x} {x^2 \left({x^2 + a^2}\right)^{n - 1} }       | c = substituting for $z$ and $\mathrm d z$ }} {{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^{n - 1} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} }       | r = \int \frac {\paren {x^2 + a^2} \rd x} {x^m \paren {x^2 + a^2}^{n - 1} \paren {x^2 + a^2} }       | c =  }} {{eqn | r = \int \frac {\paren {x^2 + a^2} \rd x} {x^m \paren {x^2 + a^2}^{\left({n - 1}\right) + 1} }       | c =  }} {{eqn | r = \int \frac {x^2 \rd x} {x^m \paren {x^2 + a^2}^n} + a^2 \int \frac {\d x} {x^m \paren {x^2 + a^2}^n}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n} + a^2 \int \frac {\d x} {x^m \paren {x^2 + a^2}^n}       | c = simplifying }} {{eqn | ll= \leadsto       | l = a^2 \int \frac {\d x} {x^m \paren {x^2 + a^2}^n}        | r = \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} } - \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}       | c = changing sides }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {x^m \paren {x^2 + a^2}^n}        | r = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 - a^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {x \paren {x^2 - a^2} }       | r = \int \frac {\d z} {2 z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\d z} z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \ln z + C       | c = Primitive of Reciprocal: Corollary as $z > 0$ }} {{eqn | r = \frac 1 2 \map \ln {x^2 - a^2} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = \int \frac {x^3 \ \mathrm d x} {x^2 - a^2}       | r = \int \frac {x \left({x^2 - a^2 + a^2}\right)} {x^2 - a^2} \ \mathrm d x       | c =  }} {{eqn | r = \int \frac {x \left({x^2 - a^2}\right)} {x^2 - a^2} \ \mathrm d x + \int \frac {a^2 x} {x^2 - a^2} \ \mathrm d x       | c =  }} {{eqn | r = \int x \ \mathrm d x + a^2 \int \frac {x \ \mathrm d x} {x^2 - a^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac {x^2} 2 + a^2 \int \frac {x \ \mathrm d x} {x^2 - a^2} + C       | c = Primitive of Power }} {{eqn | r = \frac {x^2} 2 + a^2 \left({\frac 1 2 \ln \left({x^2 - a^2}\right)}\right) + C       | c = Primitive of $\dfrac x {x^2 - a^2}$ }} {{eqn | r = \frac {x^2} 2 + \frac {a^2} 2 \ln \left({x^2 - a^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2} }       | r = \int \paren {\frac x {a^2 \paren {x^2 - a^2} }  - \frac 1 {a^2 x} } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} x       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \ln \size x + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^2} \paren {\frac 1 2 \map \ln {x^2 - a^2} } - \frac 1 {a^2} \ln \size x + C       | c = Primitive of $\dfrac x {x^2 - a^2}$ }} {{eqn | r = \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \ln \size {x^2} + C       | c = Logarithm of Power }} {{eqn | r = \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \map \ln {x^2} + C       | c = as $x^2 > 0$ }} {{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 \left({x^2 - a^2}\right)}       | r = \int \left({\frac 1 {a^2 \left({x^2 - a^2}\right)}  - \frac 1 {a^2 x^2} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\mathrm d x} {x^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^2 - a^2} + \frac 1 {a^2 x} + C       | c = Primitive of Power }} {{eqn | r = \frac 1 {a^2} \left({\frac 1 2 \ln \left ({\frac {x - a} {x + a} }\right) }\right) + \frac 1 {a^2 x} + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$ }} {{eqn | r = \frac 1 {a^2 x} + \frac 1 {2 a^3} \ln \left({\frac {x - a} {x + a} }\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^3 \left({x^2 - a^2}\right)}       | r = \int \left({\frac {-1} {a^2 x^3} - \frac 1 {a^4 x} + \frac x {a^4 \left({x^2 - a^2}\right)} }\right) \ \mathrm d x       | c = Partial Fraction Expansion }} {{eqn | r = \frac {-1} {a^2} \int \frac {\mathrm d x} {x^3} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 - a^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 - a^2} + C       | c = Primitive of Power }} {{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 - a^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \left({\frac 1 2 \ln \left ({x^2 -a^2}\right)}\right) + C       | c = Primitive of $\dfrac x {x^2 - a^2}$ }} {{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \ln \left({x^2}\right) + \frac 1 {2 a^4} \ln \left ({x^2 - a^2}\right) + C       | c = Logarithm of Power }} {{eqn | r = \frac 1 {2 a^2 x^2} + \frac 1 {2 a^4} \ln \left({\frac {x^2 - a^2} {x^2} }\right) + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {\left({x^2 - a^2}\right)^2}       | r = \int \left({\frac 1 {4 a^3 \left({x + a}\right)} - \frac 1 {4 a^3 \left({x - a}\right)} + \frac 1 {4 a^2 \left({x + a}\right)^2} + \frac 1 {4 a^2 \left({x - a}\right)^2} }\right) \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {4 a^3} \int \frac {\d x} {x + a} - \frac 1 {4 a^3} \int \frac {\d x} {x - a} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {4 a^3} \ln \left\vert{x + a}\right\vert - \frac 1 {4 a^3} \ln \left\vert{x - a}\right\vert + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2} + C       | c = Primitive of Function of $a x + b$ and Primitive of Reciprocal }} {{eqn | r = \frac 1 {4 a^3} \ln \left\vert{\frac {x + a} {x - a} }\right\vert + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2} + C       | c = Difference of Logarithms }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2} + C       | c = Sign of Quotient of Factors of Difference of Squares }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + \frac 1 {4 a^2} \frac {-1} {\left({x + a}\right)} + \frac 1 {4 a^2} \frac {-1} {\left({x - a}\right)} + C       | c = Primitive of Function of $a x + b$ and Primitive of Power }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) - \left({\frac 1 {4 a^2 \left({x + a}\right)} + \frac 1  {4 a^2 \left({x - a}\right)} }\right) + C       | c = simplifying }} {{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) - \left({\frac {\left({x - a}\right) + \left({x + a}\right)} {4 a^2 \left({x + a}\right) \left({x - a}\right)} }\right) + C       | c = simplifying }} {{eqn | r = \frac {-x} {2 a^2 \left({x^2 - a^2}\right)} + \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + C       | c = Difference of Two Squares }} {{end-eqn}} {{qed}}
We consider the conditional probability on the {{LHS}}: {{begin-eqn}} {{eqn | r = \map \Pr {X_{m + n} = j \mid X_0 = i}       | o =  }} {{eqn | r = \condprob {\paren {\bigcup_{k \mathop \in S} \sqbrk {X_{n + m} = j, X_n = k} } } {X_0 = i} }} {{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j, X_n = k} {X_0 = i}       | c = {{Defof|Countable Additivity}} }} {{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j} {X_n = k, X_0 = i} \times \condprob {X_n = k} {X_0 = i}       | c = Chain Rule for Probability }} {{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j} {X_n = k} \times \condprob {X_n = k} {X_0 = i}       | c = Markov Property }} {{eqn | r = \sum_{k \mathop \in S} {p_{i k} }^{\paren n} {p_{k j} }^{\paren m}       | c = {{Defof|Stationary Transition Probabilities}} }} {{end-eqn}} {{qed}} {{Namedfor|Sydney Chapman|name2 = Andrey Nikolaevich Kolmogorov|cat = Chapman|cat2 = Kolmogorov}} Category:Markov Chains 1hzcrk065noab0yoi3pnovz3kcvdkp8