hcju/proofwikiqa huggingface dataset:
Let: {{begin-eqn}} {{eqn | l = z       | r = a^2 - x^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 {a^2 - x^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 {a^2 - x^2} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\lambda \, \map f t + \mu \, \map g t}       | r = \int_0^{\to +\infty} e^{-s t} \paren {\lambda \, \map f t + \mu \, \map g t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \lim_{A \mathop \to +\infty} \paren {\int_0^A e^{-s t} \paren {\lambda \, \map f t + \mu \, \map g t} \rd t}       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lim_{A \mathop \to +\infty} \paren {\lambda \int_0^A e^{-s t} \map f t \rd t + \mu \int_0^A e^{-s t} \map g t \rd t}       | c = distributing $e^{-s t}$, Linear Combination of Complex Integrals }} {{eqn | r = \lambda \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map f t \rd t + \mu \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map g t \rd t       | c = Combination Theorem for Limits at Infinity }} {{eqn | r = \lambda \int_0^{\to +\infty} e^{-s t} \map f t \rd t + \mu \int_0^{\to +\infty} e^{-st} \map g t \rd t       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lambda \laptrans {\map f t} + \mu \laptrans {\map g t}       | c = {{Defof|Laplace Transform}} }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = \int \frac {x^3 \ \mathrm d x} {a^2 - x^2}       | r = \int \frac {x \left({x^2 - a^2 + a^2}\right)} {a^2 - x^2} \ \mathrm d x       | c =  }} {{eqn | r = \int \frac {-x \left({a^2 - x^2}\right)} {a^2 - x^2} \ \mathrm d x + \int \frac {a^2 x} {a^2 - x^2} \ \mathrm d x       | c =  }} {{eqn | r = -\int x \ \mathrm d x + a^2 \int \frac {x \ \mathrm d x} {a^2 - x^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -\frac {x^2} 2 + a^2 \int \frac {x \ \mathrm d x} {a^2 - x^2} + C       | c = Primitive of Power }} {{eqn | r = -\frac {x^2} 2 + a^2 \left({-\frac 1 2 \ln \left({a^2 - x^2}\right)}\right) + C       | c = Primitive of Reciprocal of $\dfrac x {a^2 - x^2}$ }} {{eqn | r = -\frac {x^2} 2 - \frac {a^2} 2 \ln \left({a^2 - x^2}\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {a^2 - x^2} }       | r = \int \paren {\frac 1 {a^2 \paren {a^2 - x^2} } + \frac 1 {a^2 x^2} } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {\d x} {x^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {a^2 - x^2} + \frac 1 {a^2 x} + C       | c = Primitive of Power }} {{eqn | r = \frac 1 {a^2} \paren {\frac 1 2 \map \ln {\frac {a + x} {a - x} } } - \frac 1 {a^2 x} + C       | c = Primitive of $\dfrac 1 {a^2 - x^2}$ }} {{eqn | r = \frac {-1} {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {a + x} {a - x} } + C       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = a^2 - x^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({a^2 - x^2}\right)^2}       | r = \int \frac {\mathrm d z} {-2 z^2}       | c = Integration by Substitution }} {{eqn | r = -\frac 1 2 \int \frac {\mathrm d z} {z^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -\frac 1 2 \left({\frac {-1} z}\right) + C       | c = Primitive of Power }} {{eqn | r = \frac 1 {2 \left({a^2 - x^2}\right)} + C       | c = substituting for $z$ and simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \rd x} {\paren {a^2 - x^2}^2}       | r = \int \frac {x^2 - a^2 + a^2} {\paren {a^2 - x^2}^2} \rd x       | c =  }} {{eqn | r = \int \frac {-\paren {a^2 - x^2} } {\paren {a^2 - x^2}^2} \rd x + a^2 \int \frac {\d x} {\paren {a^2 - x^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = -\int \frac {\d x} {a^2 - x^2} + a^2 \int \frac {\d x} {\paren {a^2 - x^2}^2}       | c = simplification }} {{eqn | r = -\frac 1 {2 a} \map \ln {\frac {a + x} {a - x} } + a^2 \int \frac {\d x} {\paren {x^2 - a^2}^2} + C       | c = Primitive of Reciprocal of $a^2 - x^2$ }} {{eqn | r = -\frac 1 {2 a} \map \ln {\frac {a + x} {a - x} } + a^2 \paren {\frac x {2 a^2 \paren {a^2 - x^2} } + \frac 1 {4 a^3} \map \ln {\frac {a + x} {a - x} } } + C       | c = Primitive of Reciprocal of $\paren {a^2 - x^2}^2$ }} {{eqn | r = \frac x {2 \paren {a^2 - x^2} } - \frac 1 {4 a} \map \ln {\frac {a + x} {a - x} } + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^3 \rd x} {\paren {a^2 - x^2}^2}       | r = \int \frac {x \paren {x^2 - a^2 + a^2} } {\paren {a^2 - x^2}^2} \rd x       | c =  }} {{eqn | r = \int \frac {-x \paren {a^2 - x^2} } {\paren {a^2 - x^2}^2} \rd x + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {-x \rd x} {a^2 - x^2} + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2}       | c = simplification }} {{eqn | r = -\frac 1 2 \paren {-\map \ln {a^2 - x^2} } + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C       | c = Primitive of $\dfrac x {a^2 - x^2}$ }} {{eqn | r = \frac 1 2 \map \ln {a^2 - x^2} + a^2 \paren {\frac 1 {2 \paren {a^2 - x^2} } } + C       | c = Primitive of Reciprocal of $\dfrac x {\paren {a^2 - x^2}^2}$ }} {{eqn | r = \frac {a^2} {2 \paren {a^2 - x^2} } + \frac 1 2 \map \ln {a^2 - x^2} + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x \paren {a^2 - x^2}^2}       | r = \int \paren {\frac 1 {a^4 x} + \frac x {a^4 \paren {a^2 - x^2} } + \frac x {a^2 \paren {a^2 - x^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} {a^2 - x^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \paren {-\frac 1 2 \map \ln {a^2 - x^2} } + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C       | c = Primitive of $\dfrac x {a^2 - x^2}$ }} {{eqn | r = \frac 1 {a^4} \ln \size x - \frac 1 {2 a^4} \map \ln {a^2 - x^2} + \frac 1 {a^2} \paren {\frac 1 {2 \paren {a^2 - x^2} } } + C       | c = Primitive of $\dfrac x {\paren {a^2 - x^2}^2}$ }} {{eqn | r = \frac 1 {2 a^4} \map \ln {x^2} - \frac 1 {2 a^4} \map \ln {a^2 - x^2} + \frac 1 {2 a^2 \paren {a^2 - x^2} } + C       | c = Logarithm of Power and $x^2 > 0$ }} {{eqn | r = \frac 1 {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {a^2 - x^2}^2}       | r = \int \paren {\frac 1 {a^4 x^2} + \frac 3 {4 a^5 \paren {a + x} } + \frac 3 {4 a^5 \paren {a - x} } + \frac 1 {4 a^4 \paren {a + x}^2} + \frac 1 {4 a^4 \paren {a - x}^2} } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} + \frac 3 {4 a^5} \int \frac {\d x} {a + x} + \frac 3 {4 a^5} \int \frac {\d x} {a - x} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \frac {-1} x + \frac 3 {4 a^5} \int \frac {\d x} {a + x} + \frac 3 {4 a^5} \int \frac {\d x} {a - x} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {a + x} + \frac 3 {4 a^5} \paren {-\ln \size {a - x} } + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {a + x} - \frac 3 {4 a^5} \ln \size {a - x} + \frac 1 {4 a^4} \frac {-1} {a + x} + \frac 1 {4 a^4} \frac 1 {a - x} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {\frac {a + x} {a - x} } + \frac {-1} {4 a^4 \paren {a + x} } + \frac 1 {4 a^4 \paren {a - x} } + C       | c = Difference of Logarithms }} {{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + \frac {-1} {4 a^4 \paren {a + x} } + \frac 1 {4 a^4 \paren {a - x} } + C       | c = Sign of Quotient of Factors of Difference of Squares }} {{eqn | r = \frac {-1} {a^4 x} + \frac x {2 a^4 \paren {a^2 - x^2} } + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + 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({a^2 - x^2}\right)^{n - 1} }       | c =  }} {{eqn | r = \left({a^2 - x^2}\right)^{- \left({n - 1}\right)}       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = - \left({n - 1}\right) \left({-2 x}\right) \left({a^2 - x^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({a^2 - x^2}\right)^{-n}       | c =  }} {{eqn | r = \frac {\left({2 n - 2}\right) x} {\left({a^2 - x^2}\right)^n}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} }       | r = \frac 1 {\left({a^2 - x^2}\right)^{n - 1} } x - \int x \frac {\left({2 n - 2}\right) x} {\left({a^2 - x^2}\right)^n} \ \mathrm d x       | c = Integration by Parts }} {{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) \int \frac {x^2} {\left({a^2 - x^2}\right)^n} \ \mathrm d x       | c = simplifying }} {{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) \int \frac {x^2 - a^2 + a^2} {\left({a^2 - x^2}\right)^n} \ \mathrm d x       | c =  }} {{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) \int \frac {-\left({a^2 - x^2}\right)} {\left({a^2 - x^2}\right)^n} \ \mathrm d x - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n}       | c = Linear Combination of Integrals }} {{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n}        | c =  }} {{eqn | ll= \implies       | l = \left({1 - \left({2 n - 2}\right)}\right) \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} }       | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n}       | c =  }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n}       | r = \frac x {2 \left({n - 1}\right) a^2 \left({a^2 - x^2}\right)^{n - 1} } + \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} }       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = a^2 - x^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 {x \rd x} {\paren {a^2 - x^2}^n}       | r = \int \frac {\d z} {- 2 z^n}       | c = Integration by Substitution }} {{eqn | r = \frac {-1} 2 \int z^{-n} \rd 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 \paren {n - 1} z^{n - 1} } + C       | c = simplifying }} {{eqn | r = \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} } + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = a^2 - x^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({a^2 - x^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) \left({-a^2}\right)} - \frac {\left({-n + \left({-1}\right) + 2}\right)} {\left({-n + 1}\right) \left({-a^2}\right)} \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({a^2 - x^2}\right)^{n - 1} } + \frac 1 {2 a^2} \int \frac {2 x \ \mathrm d x} {x^2 \left({a^2 - x^2}\right)^{n - 1} }       | c = substituting for $z$ and $\mathrm d z$ }} {{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 \left({a^2 - x^2}\right)^{n - 1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)^{n - 1} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^m \rd x} {\paren {a^2 - x^2}^n}       | r = \int \frac {x^{m - 2} \paren {x^2} \rd x} {\paren {a^2 - x^2}^n}       | c =  }} {{eqn | r = \int \frac {x^{m - 2} \paren {a^2 + x^2 - a^2} \rd x} {\paren {a^2 - x^2}^n}       | c =  }} {{eqn | r = \int \frac {x^{m - 2} \paren {a^2 - \paren {a^2 - x^2} } \rd x} {\paren {a^2 - x^2}^n}       | c =  }} {{eqn | r = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \paren {a^2 - x^2} \rd x} {\paren {a^2 - x^2}^n}       | c = Linear Combination of Integrals }} {{eqn | r = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^{n - 1} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^{n - 1} }       | r = \int \frac {\left({a^2 - x^2}\right) \ \mathrm d x} {x^m \left({a^2 - x^2}\right)^{n - 1} \left({a^2 - x^2}\right)}       | c =  }} {{eqn | r = \int \frac {\left({a^2 - x^2}\right) \ \mathrm d x} {x^m \left({a^2 - x^2}\right)^{\left({n - 1}\right) + 1} }       | c =  }} {{eqn | r = a^2 \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n} - \int \frac {x^2 \ \mathrm d x} {x^m \left({a^2 - x^2}\right)^n}       | c = Linear Combination of Integrals }} {{eqn | r = a^2 \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n} - \int \frac {\mathrm d x} {x^{m - 2} \left({a^2 - x^2}\right)^n}       | c = simplifying }} {{eqn | ll= \implies       | l = a^2 \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n}        | r = \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^{n - 1} } + \int \frac {\mathrm d x} {x^{m - 2} \left({a^2 - x^2}\right)^n}       | c = changing sides }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n}        | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^{n-1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x^{m-2} \left({a^2 - x^2}\right)^n}       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z^2       | r = x^2 + a^2       | c =  }} {{eqn | ll= \leadsto       | l = 2 z \frac {\d z} {\d x}       | r = 2 x       | c = Chain Rule for Derivatives, Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \int \frac {x \rd x} {\sqrt {x^2 + a^2} }       | r = \int \frac {z \rd z} {z}       | c = Integration by Substitution }} {{eqn | r = \int \rd z       | c =  }} {{eqn | r = z + C       | c = Primitive of Constant }} {{eqn | r = \sqrt {x^2 + a^2} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
With a view to expressing the problem in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = x^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = \frac x {\sqrt {x^2 + a^2} }       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \sqrt {x^2 + a^2}       | c = Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x^3 \rd x} {\sqrt {x^2 + a^2} }       | r = \int x^2 \frac {x \rd x} {\sqrt {x^2 + a^2} }       | c =  }} {{eqn | r = x^2 \sqrt {x^2 + a^2} - \int 2 x \sqrt {x^2 + a^2} \rd x       | c = Integration by Parts }} {{eqn | r = x^2 \sqrt {x^2 + a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 + a^2} }^3} 3 } + C       | c = Primitive of $x \sqrt {x^2 + a^2}$ }} {{eqn | r = \paren {x^2 + a^2 - a^2} \sqrt {x^2 + a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 + a^2} }^3} 3 } + C       | c =  }} {{eqn | r = \paren {\sqrt {x^2 + a^2} }^3 - a^2 \sqrt {x^2 + a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 + a^2} }^3} 3 } + C       | c =  }} {{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 - a^2 \sqrt {x^2 + a^2} + C       | c =  }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 }} {{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^2 \sqrt {x^2 + a^2} }       | r = \int \frac {\mathrm d z} {2 z \sqrt z \sqrt {z + a^2} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z + a^2} }       | c =  }} {{end-eqn}} Using Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$: :$\displaystyle \int \frac {\mathrm d x} {x^m \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 3}\right) a} {\left({2 m - 2}\right) b} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$ Setting: {{begin-eqn}} {{eqn | l = x       | o = :=       | r = z }} {{eqn | l = m       | o = :=       | r = \frac 3 2 }} {{eqn | l = a       | o = :=       | r = 1 }} {{eqn | l = b       | o = :=       | r = a^2 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z + a^2} }       | r = \frac {-\sqrt{z + a^2} } {2 \left({\left({\frac 3 2}\right) - 1}\right) a^2 z^{\left({3/2}\right) - 1} } - \frac {2 \left({\frac 3 2}\right) - 3} {2 \left({2 \left({\frac 3 2}\right) - 2}\right) a^2} \int \frac {\mathrm d z} {z^{\left({\frac 3 2}\right) - 1} \sqrt{z + a^2} } + C       | c = Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$ }} {{eqn | r = \frac {-\sqrt{z + a^2} } {a^2 z^{1/2} } - 0 + C       | c = simplifying }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } {a^2 x} + C       | c = substituting back for $z$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 }} {{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^3 \sqrt {x^2 + a^2} }       | r = \int \frac {\mathrm d z} {2 z^{3/2} \sqrt z \sqrt {z + a^2} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {z^2 \sqrt {z + a^2} }       | c =  }} {{end-eqn}} Using Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$: :$\displaystyle \int \frac {\mathrm d x} {x^m \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 3}\right) a} {\left({2 m - 2}\right) b} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$ Setting: {{begin-eqn}} {{eqn | l = x       | o = :=       | r = z }} {{eqn | l = m       | o = :=       | r = 2 }} {{eqn | l = a       | o = :=       | r = 1 }} {{eqn | l = b       | o = :=       | r = a^2 }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \frac 1 2 \int \frac {\mathrm d z} {z^2 \sqrt {z + a^2} }       | r = \frac {-\sqrt{z + a^2} } {2 \left({2 - 1}\right) a^2 z^{2 - 1} } - \frac {2 \left({2}\right) - 3} {2 \left({2 \left({2}\right) - 2}\right) a^2} \int \frac {\mathrm d z} {z^{2 - 1} \sqrt{z + a^2} } + C       | c = Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$ }} {{eqn | r = \frac {-\sqrt{z + a^2} } {2 a^2 z} - \frac 1 {4 a^2} \int \frac {\mathrm d z} {z \sqrt{z + a^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\sqrt{x^2 + a^2} } {2 a^2 x^2} - \frac 1 {4 a^2} \left({\int \frac {2 x \ \mathrm d x} {x^2 \sqrt{x^2 + a^2} } }\right) + C       | c = substituting for $z$ }} {{eqn | r = \frac {-\sqrt{x^2 + a^2} } {2 a^2 x^2} - \frac 1 {2 a^2} \left({\int \frac {\mathrm d x} {x \sqrt{x^2 + a^2} } }\right) + C       | c = simplification }} {{eqn | r =\frac {- \sqrt {x^2 + a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C       | c = Primitive of $\dfrac 1 {x \sqrt{x^2 + a^2} }$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \int x \sqrt {x^2 + a^2} \rd x       | r = \int \frac {\sqrt z \sqrt {z + a^2} \rd z} {2 \sqrt z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \sqrt {z + a^2} \rd z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \frac {2 \paren {\sqrt {z + a^2} }^3} 3 + C       | c = Primitive of $\sqrt {a x + b}$ }} {{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + C       | c = substituting for $z$ and simplifying }} {{end-eqn}} {{qed}}