hcju/proofwikiqa huggingface dataset:
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^2 \sqrt {x^2 + a^2} \rd x       | r = \int \frac {z \sqrt {z + a^2} \rd z} {2 \sqrt z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \sqrt z \sqrt {z + a^2} \rd z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \paren {\frac {2 z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C       | c = Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$ }} {{eqn | r = \frac {2 z + a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} {16} \paren {2 \map \ln {\sqrt {z + a^2} + \sqrt z} } + C       | c = Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$ }} {{eqn | r = \frac {2 z + 2 a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C       | c =  }} {{eqn | r = \frac {z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C       | c =  }} {{eqn | r = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 - \frac {a^2 x \sqrt {x^2 + a^2} } 8 - \frac {a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C       | c = substituting for $z$ }} {{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^3 \sqrt {x^2 + a^2} \rd x       | r = \int \frac {z^{3/2} \sqrt {z + a^2} \rd z} {2 \sqrt z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int z \sqrt {z + a^2} \rd z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \paren {\frac {2 \paren {3 z - 2 a^2} } {15} \paren {\sqrt {z + a^2} }^3} + C       | c = Primitive of $x \sqrt {\paren {a x + b} }$ }} {{eqn | r = \frac {3 z - 2 a^2} {15} \paren {\sqrt {z + a^2} }^3 + C       | c = simplification }} {{eqn | r = \frac {3 x^2 - 2 a^2} {15} \paren {\sqrt {x^2 + a^2} }^3 + C       | c = substituting for $z$ }} {{eqn | r = \frac {3 x^2 + 3 a^2 - 5 a^2} {15} \paren {\sqrt {x^2 + a^2} }^3 + C       | c =  }} {{eqn | r = \frac {3 \paren {x^2 + a^2} \paren {\sqrt {a^2 - x^2} }^3 - 5 a^2 \paren {\sqrt {a^2 - x^2} }^3} {15} + C       | c =  }} {{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^5} 5 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^3} 3 + C       | c = simplification }} {{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 {\sqrt {x^2 + a^2} } x \ \mathrm d x       | r = \int \frac {\sqrt {z + a^2} \ \mathrm d z} {2 \sqrt z \sqrt z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\sqrt {z + a^2} \ \mathrm d z} z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \left({2 \sqrt {z + a^2} + a^2 \int \frac {\mathrm d z} {z \sqrt {z + a^2} } }\right) + C       | c = Primitive of $\dfrac {\sqrt {a x + b} } x$ }} {{eqn | r = \sqrt {x^2 + a^2} + \frac {a^2} 2 \int \frac {2 x \ \mathrm d x} {x^2 \sqrt {x^2 + a^2} } + C       | c = substituting for $z$ }} {{eqn | r = \sqrt {x^2 + a^2} + a^2 \int \frac {\mathrm d x} {x \sqrt {x^2 + a^2} } + C       | c = simplifying }} {{eqn | r = \sqrt {x^2 + a^2} + a^2 \left({-\frac 1 a \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right)}\right) + C       | c = Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$ }} {{eqn | r = \sqrt {x^2 + a^2} - a \ln \left({\frac {a + \sqrt {x^2 + a^2} } a}\right) + C       | c = simplification }} {{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 \frac {\sqrt {x^2 + a^2} } {x^2} \rd x       | r = \int \frac {\sqrt {z + a^2} \rd z} {2 z \sqrt z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\sqrt {z + a^2} \rd z} {z^{3/2} }       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \paren {\frac {-\sqrt {z + a^2} } {\frac 1 2 \sqrt z} + \frac 1 2 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C       | c = Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$ }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } x + \frac 1 2 \int \frac {2 x \rd x} {x \sqrt {x^2 + a^2} } + C       | c = substituting for $z$ }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } x + \int \frac {\d x} {\sqrt {x^2 + a^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } x + \map \ln {x + \sqrt {x^2 + a^2} } + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ }} {{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 {\sqrt {x^2 + a^2} } {x^3} \ \mathrm d x       | r = \int \frac {\sqrt {z + a^2} \ \mathrm d z} {2 z^{3/2} \sqrt z}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\sqrt {z + a^2} \ \mathrm d z} {z^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \left({\frac {-\sqrt {z + a^2} } z + \frac 1 2 \int \frac {\mathrm d z} {z \sqrt {z + a^2} } }\right) + C       | c = Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$ }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} + \frac 1 4 \int \frac {2 x \ \mathrm d x} {x^2 \sqrt {x^2 + a^2} } + C       | c = substituting for $z$ }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} + \frac 1 2 \int \frac {\mathrm d x} {x \sqrt {x^2 + a^2} } + C       | c = simplifying }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} + \frac 1 2 \left({-\frac 1 a \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) }\right) + C       | c = Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$ }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} - \frac 1 {2 a} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C       | c = simplifying }} {{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 \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}       | r = \int \frac {\d z} {2 \sqrt z \paren {\sqrt {z + a^2} }^3}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\d z} {\paren {z + a^2} \sqrt z \sqrt {z + a^2} }       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \paren {\frac {2 \sqrt z} {a^2 \sqrt {z + a^2} } } + C       | c = Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$ }} {{eqn | r = \frac 1 2 \paren {\frac {2 x} {a^2 \sqrt {x^2 + a^2} } } + C       | c = substituting for $z$ }} {{eqn | r = \frac x {a^2 \sqrt {x^2 + a^2} } + C       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 + a^2 }} {{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 {\sqrt {x^2 + a^2} }^3}       | r = \int \frac {x \rd z} {2 x z^{3/2} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int z^{-3/2} \rd z       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \paren {\frac {-z^{1/2} } {\frac 1 2} } + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {\sqrt z} + C       | c = simplifying }} {{eqn | r = \frac {-1} {\sqrt {x^2 + a^2} } + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
With a view to expressing the problem in the form: :$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \mathrm d x$ let: {{begin-eqn}} {{eqn | l = u       | r = x }} {{eqn | ll= \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = 1       | c = Power Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\mathrm d v} {\mathrm d x}       | r = \frac x {\left({\sqrt {x^2 + a^2} }\right)^3} }} {{eqn | ll= \implies       | l = v       | r = \frac {-1} {\sqrt {x^2 + a^2} }       | c = Primitive of $\dfrac x {\left({\sqrt {x^2 + a^2} }\right)^3}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x^2 \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}       | r = \int x \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}       | c =  }} {{eqn | r = x \frac {-1} {\sqrt {x^2 + a^2} } - \int \frac {-1} {\sqrt {x^2 + a^2} } 1 + C       | c = Integration by Parts }} {{eqn | r = \frac {-x} {\sqrt {x^2 + a^2} } + \int \frac 1 {\sqrt {x^2 + a^2} } + C       | c = simplifying }} {{eqn | r = \frac {-x} {\sqrt {x^2 + a^2} } + \ln \left({x + \sqrt {x^2 + a^2} }\right) + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^3 \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}       | r = \int \frac {x \left({x^2}\right) \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}       | c =  }} {{eqn | r = \int \frac {x \left({x^2 + a^2 - a^2}\right) \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}       | c =  }} {{eqn | r = \int \frac {x \left({x^2 + a^2}\right) \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3} - a^2 \int \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {x \ \mathrm d x} {\sqrt {x^2 + a^2} } - a^2 \int \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}       | c = simplifying }} {{eqn | r = \sqrt {x^2 + a^2} - a^2 \int \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3} + C       | c = Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$ }} {{eqn | r = \sqrt {x^2 + a^2} - a^2 \frac {-1} {\sqrt {x^2 + a^2} } + C       | c = Primitive of $\dfrac x {\left({\sqrt {x^2 + a^2} }\right)^3}$ }} {{eqn | r = \sqrt {x^2 + a^2} + \frac {a^2} {\sqrt {x^2 + a^2} } + C       | c = simplifying }} {{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 \frac {\d x} {x \paren {\sqrt {x^2 + a^2} }^3}       | r = \int \frac {\d z} {2 \sqrt z \sqrt z \paren {\sqrt {z + a^2} }^3}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \int \frac {\d z} {z \paren {\sqrt {z + a^2} }^3}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 2 \paren {\frac 2 {a^2 \sqrt {z + a^2} } + \frac 1 {a^2} \int \frac {\d z} {z \sqrt {z + a^2} } } + C       | c = Primitive of $\dfrac 1 {x \paren {\sqrt{a x + b} }^m }$ }} {{eqn | r = \frac 1 {a^2 \sqrt {z + a^2} } + \frac 1 {2 a^2} \int \frac {\d z} {z \sqrt {z + a^2} } + C       | c = simplifying }} {{eqn | r = \frac 1 {a^2 \sqrt {x^2 + a^2} } + \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \sqrt {x^2 + a^2} } + C       | c = substituting for $z$ }} {{eqn | r = \frac 1 {a^2 \sqrt {x^2 + a^2} } + \frac 1 {a^2} \int \frac {\d x} {x \sqrt {x^2 + a^2} } + C       | c = simplifying }} {{eqn | r = \frac 1 {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C       | c = Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {\sqrt {x^2 + a^2} }^3}       | r = \int \frac {a^2 \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3}       | c =  }} {{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3}       | c =  }} {{eqn | r = \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3} - \frac 1 {a^2} \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {x^2 + a^2} }^3}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 \sqrt {x^2 + a^2} } - \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}       | c = simplifying }} {{eqn | r = \frac 1 {a^2} \frac {-\sqrt {x^2 + a^2} } {a^2 x} - \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3} + C       | c = Primitive of $\dfrac 1 {x^2 \sqrt {x^2 + a^2} }$ }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac 1 {a^2} \frac x {a^2 \sqrt {x^2 + a^2} } + C       | c = Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$ }} {{eqn | r = \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 + a^2} } + C       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^3 \left({\sqrt {x^2 + a^2} }\right)^3}       | r = \int \frac {a^2 \ \mathrm d x} {a^2 x^3 \left({\sqrt {x^2 + a^2} }\right)^3}       | c =  }} {{eqn | r = \int \frac {\left({x^2 + a^2 - x^2}\right) \ \mathrm d x} {a^2 x^3 \left({\sqrt {x^2 + a^2} }\right)^3}       | c =  }} {{eqn | r = \frac 1 {a^2} \int \frac {\left({x^2 + a^2}\right) \ \mathrm d x} {x^3 \left({\sqrt {x^2 + a^2} }\right)^3} - \frac 1 {a^2} \int \frac {x^2 \ \mathrm d x} {x^3 \left({\sqrt {x^2 + a^2} }\right)^3}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^3 \sqrt {x^2 + a^2} } - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {x^2 + a^2} }\right)^3}       | c = simplifying }} {{eqn | r = \frac 1 {a^2} \left({\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)}\right) - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {x^2 + a^2} }\right)^3} + C       | c = Primitive of $\dfrac 1 {x^3 \sqrt {x^2 + a^2} }$ }} {{eqn | r = \frac 1 {a^2} \left({\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)}\right)       | c =  }} {{eqn | o =        | ro= -       | r = \frac 1 {a^2} \left({\frac 1 {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) }\right) + C       | c = Primitive of $\dfrac 1 {x \left({\sqrt {x^2 + a^2} }\right)^3}$ }} {{eqn | r = \frac {-1} {2 a^2 x^2 \sqrt {x^2 + a^2} } - \frac 3 {2 a^4 \sqrt {x^2 + a^2} } + \frac 3 {2 a^5} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C       | c = simplification }} {{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 \paren {\sqrt {x^2 + a^2} }^3 \rd x       | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 \sqrt z} \rd x       | c = Integration by Substitution }} {{eqn | r = \frac {\sqrt z \paren {\sqrt {z + a^2} }^3} 4 + \frac {3 a^2} 8 \int \frac {\sqrt {z + a^2} } {\sqrt z} \rd x + C       | c = Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt{a x + b} }$ }} {{eqn | r = \frac {\sqrt z \paren {\sqrt {z + a^2} }^3} 4 + \frac {3 a^2} 8 \paren {\sqrt z \sqrt {z + a^2} + \frac {a^2} 2 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C       | c = Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$ }} {{eqn | r = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 + \frac {3 a^2} 8 x \sqrt {x^2 + a^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {x^2 + a^2} } + C       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 + \frac {3 a^2 x \sqrt {x^2 + a^2} } 8 + \frac {3 a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 + a^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 x \left({\sqrt {x^2 + a^2} }\right)^3 \ \mathrm d x       | r = \int \frac {z^{3/2} } 2 \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \frac {z^{5/2} } {\frac 5 2} + C       | c = Primitive of Power }} {{eqn | r = \frac {z^{5/2} } 5 + C       | c = simplifying }} {{eqn | r = \frac {\left({\sqrt {x^2 + a^2} }\right)^5} 5 + C       | c = substituting 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 x^2 \left({\sqrt {x^2 + a^2} }\right)^3 \ \mathrm d x       | r = \int \frac {\sqrt z \left({z + a^2}\right)^{3/2} } 2 \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac {\sqrt z \left({z + a^2}\right)^{5/2} } 6 - \frac {a^2} {12} \int \frac {\left({z + a^2}\right)^{3/2} } {\sqrt z} \ \mathrm d z + C       | c = Primitive of $\left({p x + q}\right)^n \sqrt{a x + b}$ }} {{eqn | r = \frac {x \left({x^2 + a^2}\right)^{5/2} } 6 - \frac {a^2} 6 \int \left({x^2 + a^2}\right)^{3/2} \ \mathrm d x + C       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {x \left({x^2 + a^2}\right)^{5/2} } 6 }} {{eqn | o =       | ro= -       | r = \frac {a^2} 6 \left({\frac {x \left({\sqrt {x^2 + a^2} }\right)^3} 4 + \frac {3 a^2 x \sqrt {x^2 + a^2} } 8 + \frac {3 a^4} 8 \ln \left({x + \sqrt {x^2 + a^2} }\right)}\right) + C       | c = Primitive of $\left({\sqrt {x^2 + a^2} }\right)^3$ }} {{eqn | r = \frac {x \left({x^2 + a^2}\right)^{5/2} } 6 - \frac {a^2 x \left({\sqrt {x^2 + a^2} }\right)^3} {24} }} {{eqn | o =       | ro= -       | r = \frac {a^4 x \sqrt {x^2 + a^2} } {16} - \frac {a^6} {16} \ln \left({x + \sqrt {x^2 + a^2} }\right) + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int x^3 \paren {\sqrt {x^2 + a^2} }^3 \rd x       | r = \int x \paren {x^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x       | c =  }} {{eqn | r = \int x \paren {x^2 + a^2 - a^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x       | c = Primitive of Power }} {{eqn | r = \int x \paren {x^2 + a^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x       | c = Linear Combination of Integrals }} {{eqn | r = \int x \paren {\sqrt {x^2 + a^2} }^5 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x       | c = simplifying }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = x^2 + a^2 }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | o =        | r = \int x \paren {\sqrt {x^2 + a^2} }^5 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x       | c =  }} {{eqn | r = \int \frac {\sqrt z z^{5/2} } {2 \sqrt z} \rd z - a^2 \int \frac {\sqrt z z^{3/2} } {2 \sqrt z} \rd z       | c = Integration by Substitution }} {{eqn | r =  \frac 1 2 \int z^{5/2} \rd z - \frac {a^2} 2 \int z^{3/2} \rd z       | c = simplifying }} {{eqn | r = \frac 1 2 \frac {z^{7/2} } {7/2} - \frac {a^2} 2 \frac {z^{5/2} } {5/2} + C       | c = Primitive of Power }} {{eqn | r = \frac {z^{7/2} } 7 - a^2 \frac {z^{5/2} } 5 + C       | c = simplifying }} {{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^7} 7 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^5} 5 + C       | c = substituting for $z$ }} {{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 \frac {\paren {\sqrt {x^2 + a^2} }^3} x \rd x       | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z} \rd z       | c = Integration by Substitution }} {{eqn | r = \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \int \frac {\sqrt {z + a^2} } z \rd z       | c = Primitive of $\dfrac {\paren {\sqrt {a x + b} }^m} x$ }} {{eqn | r = \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \paren {2 \sqrt {z + a^2} + a^2 \int \frac {\d z} {z \sqrt {z + a^2} } }       | c = Primitive of $\dfrac {\sqrt {a x + b} } x$ }} {{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} + a^4 \int \frac {\d x} {x \sqrt {x^2 + a^2} }       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} - a^3 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + 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= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \implies       | l = \int \frac{\left({\sqrt {x^2 + a^2} }\right)^3} {x^2} \ \mathrm d x       | r = \int \frac{\left({\sqrt {z + a^2} }\right)^3} {2 z \sqrt z} \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac {-\left({\sqrt {z + a^2} }\right)^3} {\sqrt z} + \frac 3 2 \int \frac{\sqrt {z + a^2} } {\sqrt z} \ \mathrm d z       | c = Primitive of $\dfrac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n}$: Formulation 3 }} {{eqn | r = \frac {-\left({\sqrt {x^2 + a^2} }\right)^3} x + 3 \int \sqrt {x^2 + a^2} \ \mathrm d x       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {-\left({\sqrt {x^2 + a^2} }\right)^3} x + \frac{3 x \sqrt {x^2 + a^2} } 2 + \frac {3 a^2} 2 \ln \left({x + \sqrt {x^2 + a^2} }\right) + C       | c = Primitive of $\sqrt {x^2 + a^2}$ }} {{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 \frac {\paren {\sqrt {x^2 + a^2} }^3} {x^3} \rd x       | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z^2} \rd z       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \paren {\frac {-\paren {\sqrt {z + a^2} }^3} z + \frac 3 2 \int \frac {\sqrt {z + a^2} } z \rd z}       | c = Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3 }} {{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \int \frac {\sqrt {x^2 + a^2} } x \rd x       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \paren {\sqrt {x^2 + a^2} - a \map \ln {\frac {a + \sqrt {x^2 + a^2} } a} } + C       | c = Primitive of $\dfrac {\sqrt {x^2 + a^2} } x$ }} {{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 + a^2} } 2 - \frac {3 a} 2 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C       | c = simplifying }} {{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}}