hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 \left({x^3 + a^3}\right)^2}       | r = \int \frac {a^3 \ \mathrm d x} {a^3 x^2 \left({x^3 + a^3}\right)^2}       | c = multiplying top and bottom by $a^3$ }} {{eqn | r = \int \frac {\left({x^3 + a^3 - x^3}\right) \ \mathrm d x} {a^3 x^2 \left({x^3 + a^3}\right)^2}       | c =  }} {{eqn | r = \frac 1 {a^3} \int \frac {\left({x^3 + a^3}\right) \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)^2} - \frac 1 {a^3} \int \frac {x^3 \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^3} \int \frac {\mathrm d x} {x^2 \left({x^3 + a^3}\right)} - \frac 1 {a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | c = simplifying }} {{eqn | r = \frac 1 {a^3} \left({\frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)} }\right) - \frac 1 {a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | c = Primitive of $\dfrac 1 {x^2 \left({x^3 + a^3}\right)}$: Lemma }} {{eqn | r = \frac {-1} {a^6 x} - \frac 1 {a^6} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)} - \frac 1 {a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | c = simplifying }} {{end-eqn}} Now consider: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | r = \int \frac {x^3 \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)^2}       | c = multiplying top and bottom by $x^2$ }} {{eqn | r = \int \frac {\left({x^3 + a^3 - a^3}\right) \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)^2}       | c =  }} {{eqn | r = \int \frac {\left({x^3 + a^3}\right) \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)^2} - \int \frac {a^3 \ \mathrm d x} {x^2 \left({x^3 + a^3}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {\mathrm d x} {x^2 \left({x^3 + a^3}\right)} - a^3 \int \frac {\mathrm d x} {x^2 \left({x^3 + a^3}\right)^2}       | c = simplifying }} {{end-eqn}} {{proof wanted|seems to be going nowhere}}
{{begin-eqn}} {{eqn | l = \int \frac {x^m \rd x} {x^3 + a^3}       | r = \int \frac {x^{m - 3} \left({x^3}\right) \rd x} {x^3 + a^3}       | c =  }} {{eqn | r = \int \frac {x^{m - 3} \left({x^3 + a^3 - a^3}\right) \rd x} {x^3 + a^3}       | c =  }} {{eqn | r = \int \frac {x^{m - 3} \left({x^3 + a^3}\right) \rd x} {x^3 + a^3} - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}       | c = Linear Combination of Integrals }} {{eqn | r = \int x^{m - 3} \rd x - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}       | c = simplification }} {{eqn | r = \frac {x^{m - 2} } {m - 2} - a^3 \int \frac {x^{m - 3} \rd x} {x^3 + a^3}       | c = Primitive of Power }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^4 + a^4}       | r = \int \paren {\frac {x + a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 + a x \sqrt 2 + a^2} } - \frac {x - a \sqrt 2} {2 a^3 \sqrt 2 \paren {x^2 - a x \sqrt 2 + a^2} } } \rd x       | c = Partial Fraction Expansion }} {{eqn | n = 1       | r = \frac 1 {4 a^3 \sqrt 2} \int \frac {\paren {2 x + 2 a \sqrt 2} \rd x} {x^2 + a x \sqrt 2 + a^2} - \frac 1 {4 a^3 \sqrt 2} \int \frac {\paren {2 x - 2 a \sqrt 2} \rd x} {x^2 - a x \sqrt 2 + a^2}       | c = Linear Combination of Integrals }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\paren {2 x + 2 a \sqrt 2} \rd x} {x^2 + a x \sqrt 2 + a^2}       | r = \int \frac {\paren {2 x + a \sqrt 2} \rd x} {x^2 + a x \sqrt 2 + a^2} + a \sqrt 2 \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2}       | c = Linear Combination of Integrals }} {{eqn | r = \ln \size {x^2 + a x \sqrt 2 + a^2} + a \sqrt 2 \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2}       | c = Primitive of Function under its Derivative }} {{eqn | r = \ln \size {x^2 + a x \sqrt 2 + a^2} + a \sqrt 2 \paren {\frac {\sqrt 2} a \, \map \arctan {1 + \frac {x \sqrt 2} a} }       | c = Primitive of $\dfrac 1 {x^2 + a x \sqrt 2 + a^2}$ }} {{eqn | n = 2       | r = \ln \size {x^2 + a x \sqrt 2 + a^2} + 2 \map \arctan {1 + \frac {x \sqrt 2} a}       | c = simplifying }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \int \frac {\paren {2 x - 2 a \sqrt 2} \rd x} {x^2 - a x \sqrt 2 + a^2}       | r = \int \frac {\paren {2 x - a \sqrt 2} \rd x} {x^2 - a x \sqrt 2 + a^2} - a \sqrt 2 \int \frac {\d x} {x^2 - a x \sqrt 2 + a^2}       | c = Linear Combination of Integrals }} {{eqn | r = \ln \size {x^2 - a x \sqrt 2 + a^2} - a \sqrt 2 \int \frac {\d x} {x^2 - a x \sqrt 2 + a^2}       | c = Primitive of Function under its Derivative }} {{eqn | r = \ln \size {x^2 - a x \sqrt 2 + a^2} - a \sqrt 2 \paren {\frac {-\sqrt 2} a \, \map \arctan {1 - \frac {x \sqrt 2} a} }       | c = Primitive of $\dfrac 1 {x^2 - a x \sqrt 2 + a^2}$ }} {{eqn | n = 3       | r = \ln \size {x^2 - a x \sqrt 2 + a^2} + 2 \, \map \arctan {1 - \frac {x \sqrt 2} a}       | c = simplifying }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^4 + a^4}       | r = \frac 1 {4 a^3 \sqrt 2} \int \frac {\paren {2 x + 2 a \sqrt 2} \rd x} {x^2 + a x \sqrt 2 + a^2} - \frac 1 {4 a^3 \sqrt 2} \int \frac {\paren {2 x - 2 a \sqrt 2} \rd x} {x^2 - a x \sqrt 2 + a^2}       | c = from $(1)$ }} {{eqn | r = \frac 1 {4 a^3 \sqrt 2} \paren {\ln \size {x^2 + a x \sqrt 2 + a^2} + 2 \, \map \arctan {1 + \frac {x \sqrt 2} a} }        | c = from $(2)$ }} {{eqn | o =        | ro= -       | r = \frac 1 {4 a^3 \sqrt 2} \paren {\ln \size {x^2 - a x \sqrt 2 + a^2} + 2 \, \map \arctan {1 - \frac {x \sqrt 2} a} }        | c = from $(3)$ }} {{eqn | r = \frac 1 {4 a^3 \sqrt 2} \map \ln {\frac {x^2 + a x \sqrt 2 + a^2} {x^2 - a x \sqrt 2 + a^2} } - \frac 1 {2 a^3 \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = x^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = 2 x       | c =  }} {{eqn | ll= \leadsto       | l = \int \frac {x \rd x} {x^4 + a^4}       | r = \int \frac {\d z} {2 \paren {z^2 + \paren {a^2}^2} }       | c = Integration by Substitution }} {{eqn | r = \frac 1 {2 a^2} \arctan \frac z {a^2} + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \frac 1 {2 a^2} \arctan \frac {x^2} {a^2} + C       | c = substituting back for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \rd x} {x^4 + a^4}       | r = \int \paren {\frac x {2 a \sqrt 2 \paren {x^2 - a x \sqrt 2 + a^2} } - \frac x {2 a \sqrt 2 \paren {x^2 + a x \sqrt 2 + a^2} } } \rd x       | c = Partial Fraction Expansion }} {{eqn | n = 1       | r = \frac 1 {4 a \sqrt 2} \int \frac {2 x \rd x} {x^2 - a x \sqrt 2 + a^2} - \frac 1 {4 a \sqrt 2} \int \frac {2 x \rd x} {x^2 + a x \sqrt 2 + a^2}       | c = Linear Combination of Integrals }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {2 x \rd x} {x^2 - a x \sqrt 2 + a^2}       | r = \int \frac {\paren {2 x - a \sqrt 2 + a \sqrt 2} \rd x} {x^2 - a x \sqrt 2 + a^2}       | c =  }} {{eqn | r = \int \frac {\paren {2 x - a \sqrt 2} \rd x} {x^2 - a x \sqrt 2 + a^2} + a \sqrt 2 \int \frac {\d x} {x^2 - a x \sqrt 2 + a^2}       | c = Linear Combination of Integrals }} {{eqn | r = \ln \size {x^2 - a x \sqrt 2 + a^2} + a \sqrt 2 \int \frac {\d x} {x^2 - a x \sqrt 2 + a^2}       | c = Primitive of Function under its Derivative }} {{eqn | r = \ln \size {x^2 - a x \sqrt 2 + a^2} + a \sqrt 2 \paren {\frac {-\sqrt 2} a \, \map \arctan {1 - \frac {x \sqrt 2} a} }       | c = Primitive of $\dfrac 1 {x^2 - a x \sqrt 2 + a^2}$ }} {{eqn | n = 2       | r = \ln \size {x^2 - a x \sqrt 2 + a^2} - 2 \, \map \arctan {1 - \frac {x \sqrt 2} a}       | c = simplifying }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \int \frac {2 x \rd x} {x^2 + a x \sqrt 2 + a^2}       | r = \int \frac {\paren {2 x + a \sqrt 2 - a \sqrt 2} \rd x} {x^2 + a x \sqrt 2 + a^2}       | c =  }} {{eqn | r = \int \frac {\paren {2 x + a \sqrt 2} \rd x} {x^2 + a x \sqrt 2 + a^2} - a \sqrt 2 \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2}       | c = Linear Combination of Integrals }} {{eqn | r = \ln \size {x^2 + a x \sqrt 2 + a^2} - a \sqrt 2 \int \frac {\d x} {x^2 + a x \sqrt 2 + a^2}       | c = Primitive of Function under its Derivative }} {{eqn | r = \ln \size {x^2 + a x \sqrt 2 + a^2} - a \sqrt 2 \paren {\frac {\sqrt 2} a \, \map \arctan {1 + \frac {x \sqrt 2} a} }       | c = Primitive of $\dfrac 1 {x^2 + a x \sqrt 2 + a^2}$ }} {{eqn | n = 3       | r = \ln \size {x^2 + a x \sqrt 2 + a^2} - 2 \, \map \arctan {1 + \frac {x \sqrt 2} a}       | c = simplifying }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^4 + a^4}       | r = \frac 1 {4 a \sqrt 2} \int \frac {2 x \rd x} {x^2 - a x \sqrt 2 + a^2} - \frac 1 {4 a \sqrt 2} \int \frac {2 x \rd x} {x^2 + a x \sqrt 2 + a^2}       | c = from $(1)$ }} {{eqn | r = \frac 1 {4 a \sqrt 2} \paren {\ln \size {x^2 - a x \sqrt 2 + a^2} - 2 \, \map \arctan {1 - \frac {x \sqrt 2} a} }        | c = from $(2)$ }} {{eqn | o =        | ro= -       | r = \frac 1 {4 a \sqrt 2} \paren {\ln \size {x^2 + a x \sqrt 2 + a^2} - 2 \, \map \arctan {1 + \frac {x \sqrt 2} a} }        | c = from $(3)$ }} {{eqn | r = \frac 1 {4 a \sqrt 2} \map \ln {\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} } - \frac 1 {2 a \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\mathrm d} {\mathrm d x} x^4       | r = 4 x^3       | c = Primitive of Power }} {{eqn | ll= \implies       | l = \int \frac {x^3 \ \mathrm d x} {x^4 + a^4}       | r = \frac 1 4 \ln \left\vert{x^4 + a^4}\right\vert       | c = Primitive of Function under its Derivative }} {{eqn | r = \frac {\ln \left({x^4 + a^4}\right)} 4       | c = Absolute Value of Even Power }} {{end-eqn}} {{qed}}
From Primitive of Power of x less one over Power of x plus Power of a: :$\displaystyle \int \frac {x^{n - 1} \ \mathrm d x} {x^n + a^n} = \frac 1 n \ln \left\vert{x^n + a^n}\right\vert + C$ So: {{begin-eqn}} {{eqn | l = \int \frac {x^3 \ \mathrm d x} {x^4 + a^4}       | r = \frac 1 4 \ln \left\vert{x^4 + a^4}\right\vert + C       | c = Primitive of $\dfrac {x^{n - 1} } {\left({x^n + a^n}\right)}$ with $n = 4$ }} {{eqn | r = \frac {\ln \left({x^4 + a^4}\right)} 4 + C       | c = Absolute Value of Even Power }} {{end-eqn}} {{qed}}
From Primitive of Reciprocal of x by Power of x plus Power of a: :$\displaystyle \int \frac {\mathrm d x} {x \left({x^n + a^n}\right)} = \frac 1 {n a^n} \ln \left\vert{\frac {x^n} {x^n + a^n} }\right\vert + C$ So: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)}       | r = \frac 1 {4 a^4} \ln \left\vert{\frac {x^4} {x^4 + a^4} }\right\vert + C       | c = Primitive of $\dfrac 1 {x \left({x^n + a^n}\right)}$ with $n = 4$ }} {{eqn | r = \frac 1 {4 a^4} \ln \left({\frac {x^4} {x^4 + a^4} }\right) + C       | c = Absolute Value of Even Power }} {{end-eqn}} directly. {{qed}}
{{begin-eqn}} {{eqn | r = \int \frac {\mathrm d x} {x^2 \left({x^4 + a^4}\right)}       | o =        | c =  }} {{eqn | r = \int \frac {a^4 \ \mathrm d x} {a^4 x^2 \left({x^4 + a^4}\right)}       | c = multiplying top and bottom by $a^4$ }} {{eqn | r = \int \frac {\left({x^4 + a^4 - x^4}\right) \ \mathrm d x} {a^4 x^2 \left({x^4 + a^4}\right)}       | c =  }} {{eqn | r = \frac 1 {a^4} \int \frac {\left({x^4 + a^4}\right) \ \mathrm d x} {x^2 \left({x^4 + a^4}\right)} - \frac 1 {a^4} \int \frac {x^4 \ \mathrm d x} {x^2 \left({x^4 + a^4}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \int \frac {\mathrm d x} {x^2} - \frac 1 {a^4} \int \frac {x^2 \ \mathrm d x} {x^4 + a^4}       | c = simplifying }} {{eqn | r = \frac 1 {a^4} \left({\frac {-1} x}\right) - \frac 1 {a^4} \int \frac {x^2 \ \mathrm d x} {x^4 + a^4}       | c = Primitive of Power }} {{eqn | r = \frac {-1} {a^4 x} - \frac 1 {a^4} \left({\frac 1 {4 a \sqrt 2} \ln \left({\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} }\right) - \frac 1 {2 a \sqrt 2} \left({\arctan \left({1 - \frac {x \sqrt 2} a}\right) - \arctan \left({1 + \frac {x \sqrt 2} a}\right)}\right) }\right)       | c = Primitive of $\dfrac {x^2} {x^4 + a^4}$ }} {{eqn | r = \frac {-1} {a^4 x} - \frac {-1} {4 a^5 \sqrt 2} \ln \left({\frac {x^2 - a x \sqrt 2 + a^2} {x^2 + a x \sqrt 2 + a^2} }\right) + \frac 1 {2 a^5 \sqrt 2} \left({\arctan \left({1 - \frac {x \sqrt 2} a}\right) - \arctan \left({1 + \frac {x \sqrt 2} a}\right)}\right)       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^3 \left({x^4 + a^4}\right)}       | r = \int \frac {a^4 \ \mathrm d x} {a^4 x^3 \left({x^4 + a^4}\right)}       | c = multiplying top and bottom by $a^4$ }} {{eqn | r = \int \frac {\left({x^4 + a^4 - x^4}\right) \ \mathrm d x} {a^4 x^3 \left({x^4 + a^4}\right)}       | c =  }} {{eqn | r = \frac 1 {a^4} \int \frac {\left({x^4 + a^4}\right) \ \mathrm d x} {x^3 \left({x^4 + a^4}\right)} - \frac 1 {a^4} \int \frac {x^4 \ \mathrm d x} {x^3 \left({x^4 + a^4}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^4} \int \frac {\mathrm d x} {x^3} - \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^4 + a^4}       | c = simplifying }} {{eqn | r = \frac 1 {a^4} \left({\frac {-1} {2 x} }\right) - \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^4 + a^4}       | c = Primitive of Power }} {{eqn | r = \frac {-1} {2 a^4 x^2} - \frac 1 {a^4} \left({\frac 1 {2 a^2} \arctan \frac {x^2} {a^2} }\right)       | c = Primitive of $\dfrac x {x^4 + a^4}$ }} {{eqn | r = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^6} \arctan \frac {x^2} {a^2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^4 - a^4}       | r = \int \frac {\d x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Difference of Two Squares }} {{eqn | r = \int \frac {a^2 \rd x} {a^2 \paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = multiplying top and bottom by $a^2$ }} {{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 \paren {x^2 + a^2} \paren {x^2 - a^2} }       | c =  }} {{eqn | r = \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \d x} {\paren {x^2 + a^2} \paren {x^2 - a^2} } - \frac 1 {a^2} \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = simplifying }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\paren {x^2 - a^2 + a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Difference of Two Squares }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\paren {x^2 - a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \frac 1 {a^2} \int \frac {a^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c =  }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} {x^2 + a^2} - \int \frac {\d x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = simplification }} {{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} {x^2 + a^2} - \int \frac {\d x} {x^4 - a^4}       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = 2 \int \frac {\d x} {x^4 - a^4}       | r = \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} {x^2 + a^2}        | c = gathering terms }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {x^4 - a^4}       | r = \frac 1 {2 a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {2 a^2} \int \frac {\d x} {x^2 + a^2}        | c = gathering terms }} {{eqn | r = \frac 1 {4 a^3} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a^2} \int \frac {\d x} {x^2 + a^2}        | c = Primitive of $\dfrac 1 {x^2 - a^2}$ }} {{eqn | r = \frac 1 {4 a^3} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a^3} \arctan \frac x a       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = x^2       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 x       | c =  }} {{eqn | ll= \implies       | l = \int \frac {x \ \mathrm d x} {x^4 - a^4}       | r = \int \frac {\mathrm d z} {2 \left({z^2 - \left({a^2}\right)^2}\right)}       | c = Integration by Substitution }} {{eqn | r = \frac 1 2 \left({\frac 1 {2 a^2} \ln \left\vert{\frac {z - a^2} {z + a^2} }\right\vert}\right) + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$ }} {{eqn | r = \frac 1 {4 a^2} \ln \left\vert{\frac {x^2 - a^2} {x^2 + a^2} }\right\vert + C       | c = substituting back for $z$ and simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 \rd x} {x^4 - a^4}       | r = \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Difference of Two Squares }} {{eqn | r = \int \frac {\paren {x^2 + a^2 - a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c =  }} {{eqn | r = \int \frac {\paren {x^2 + a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} } + \int \frac {-a^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {-a^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = simplifying }} {{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\paren {x^2 - a^2 - x^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Difference of Two Squares }} {{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\paren {x^2 - a^2} \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = Linear Combination of Integrals }} {{eqn | o =        | ro= -       | r = \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c =  }} {{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\d x} {x^2 + a^2} - \int \frac {x^2 \rd x} {\paren {x^2 + a^2} \paren {x^2 - a^2} }       | c = simplification }} {{eqn | r = \int \frac {\d x} {x^2 - a^2} + \int \frac {\d x} {x^2 + a^2} - \int \frac {x^2 \rd x} {x^4 - a^4}       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = 2 \int \frac {x^2 \rd x} {x^4 - a^4}       | r = \int \frac {\d x} {x^2 - a^2} - \int \frac {\d x} {x^2 + a^2}        | c = gathering terms }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {x^4 - a^4}       | r = \frac 1 2 \int \frac {\d x} {x^2 - a^2} - \frac 1 2 \int \frac {\d x} {x^2 + a^2}        | c = gathering terms }} {{eqn | r = \frac 1 {4 a} \ln \size {\frac {x - a} {x + a} } - \frac 1 2 \int \frac {\d x} {x^2 + a^2}        | c = Primitive of $\dfrac 1 {x^2 - a^2}$ }} {{eqn | r = \frac 1 {4 a} \ln \size {\frac {x - a} {x + a} } - \frac 1 {2 a} \arctan \frac x a       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\mathrm d} {\mathrm d x} x^4       | r = 4 x^3       | c = Primitive of Power }} {{eqn | ll= \implies       | l = \int \frac {x^3 \ \mathrm d x} {x^4 - a^4}       | r = \frac {\ln \left\vert{x^4 - a^4}\right\vert} 4 + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
From Primitive of $\dfrac 1 {x \left({x^n - a^n}\right)}$: :$\displaystyle \int \frac {\mathrm d x} {x \left({x^n - a^n}\right)} = \frac 1 {n a^n} \ln \left\vert{\frac {x^n - a^n} {x^n} }\right\vert + C$ Setting $n = 4$: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({x^3 + a^3}\right)}       | r = \frac 1 {4 a^4} \ln \left\vert{\frac {x^4 - a^4} {x^4} }\right\vert + C       | c =  }} {{end-eqn}} directly. {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^2 \left({x^4 - a^4}\right)}       | r = \int \frac {a^4 \ \mathrm d x} {a^4 x^2 \left({x^4 - a^4}\right)}       | c = multiplying top and bottom by $a^4$ }} {{eqn | r = \int \frac {\left({x^4 + a^4 - x^4}\right) \ \mathrm d x} {a^4 x^2 \left({x^4 - a^4}\right)}       | c =  }} {{eqn | r = \int \frac {\left({x^4 - \left({x^4 - a^4}\right)}\right) \ \mathrm d x} {a^4 x^2 \left({x^4 - a^4}\right)}       | c =  }} {{eqn | r = \frac {-1} {a^4} \int \frac {\left({x^4 - a^4}\right) \ \mathrm d x} {x^2 \left({x^4 + a^4}\right)} + \frac 1 {a^4} \int \frac {x^4 \ \mathrm d x} {x^2 \left({x^4 - a^4}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac {-1} {a^4} \int \frac {\mathrm d x} {x^2} + \frac 1 {a^4} \int \frac {x^2 \ \mathrm d x} {x^4 - a^4}       | c = simplifying }} {{eqn | r = \frac {-1} {a^4} \left({\frac {-1} x}\right) + \frac 1 {a^4} \int \frac {x^2 \ \mathrm d x} {x^4 - a^4}       | c = Primitive of Power }} {{eqn | r = \frac 1 {a^4 x} + \frac 1 {a^4} \left({\frac 1 {4 a} \ln \left\vert{\frac {x - a} {x + a} }\right\vert + \frac 1 {2 a} \arctan \frac x a }\right) + C       | c = Primitive of $\dfrac {x^2} {x^4 - a^4}$ }} {{eqn | r = \frac 1 {a^4 x} + \frac 1 {4 a^5} \ln \left\vert{\frac {x - a} {x + a} }\right\vert + \frac 1 {2 a^5} \arctan \frac x a + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x^3 \left({x^4 - a^4}\right)}       | r = \int \frac {a^4 \ \mathrm d x} {a^4 x^3 \left({x^4 - a^4}\right)}       | c = multiplying top and bottom by $a^4$ }} {{eqn | r = \int \frac {\left({x^4 + a^4 - x^4}\right) \ \mathrm d x} {a^4 x^3 \left({x^4 - a^4}\right)}       | c =  }} {{eqn | r = \int \frac {\left({x^4 - \left({x^4 - a^4}\right)}\right) \ \mathrm d x} {a^4 x^3 \left({x^4 - a^4}\right)}       | c =  }} {{eqn | r = \frac {-1} {a^4} \int \frac {\left({x^4 - a^4}\right) \ \mathrm d x} {x^3 \left({x^4 - a^4}\right)} + \frac 1 {a^4} \int \frac {x^4 \ \mathrm d x} {x^3 \left({x^4 - a^4}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac {-1} {a^4} \int \frac {\mathrm d x} {x^3} + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^4 - a^4}       | c = simplifying }} {{eqn | r = \frac {-1} {a^4} \left({\frac {-1} {2 x} }\right) + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^4 - a^4}       | c = Primitive of Power }} {{eqn | r = \frac 1 {2 a^4 x^2} + \frac 1 {a^4} \left({\frac 1 {4 a^2} \ln \left\vert{\frac {x^2 - a^2} {x^2 + a^2} }\right\vert}\right) + C       | c = Primitive of $\dfrac x {x^4 - a^4}$ }} {{eqn | r = \frac 1 {2 a^4 x^2} + \frac 1 {4 a^6} \ln \left\vert{\frac {x^2 - a^2} {x^2 + a^2} }\right\vert + C       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({x^n + a^n}\right)}       | r = \int \frac {a^n \ \mathrm d x} {a^n x \left({x^2 + a^2}\right)}       | c = multiplying top and bottom by $a^n$ }} {{eqn | r = \int \frac {\left({x^n + a^n - x^n}\right) \ \mathrm d x} {a^n x \left({x^n + a^n}\right)}       | c = adding and subtracting $x^n$ }} {{eqn | r = \frac 1 {a^n} \int \frac {\left({x^n + a^n}\right) \ \mathrm d x} {x \left({x^n + a^n}\right)} - \frac 1 {a^n} \int \frac {x^n \ \mathrm d x} {x \left({x^n + a^n}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^n} \int \frac {\mathrm d x} x - \frac 1 {a^n} \int \frac {x^{n - 1} \ \mathrm d x} {x^n + a^n}       | c = simplifying }} {{eqn | r = \frac 1 {a^n} \ln \left\vert{x}\right\vert - \frac 1 {a^n} \int \frac {x^{n - 1} \ \mathrm d x} {x^n + a^n} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {a^n} \ln \left\vert{x}\right\vert - \frac 1 {a^n} \left({\frac 1 n \ln \left\vert{x^n + a^n}\right\vert }\right) + C       | c = Primitive of $\dfrac {x^{n - 1}} {x^n + a^n}$ }} {{eqn | r = \frac 1 {n a^n} \ln \left\vert{x^n}\right\vert - \frac 1 {n a^n} \ln \left\vert{x^n + a^n}\right\vert + C       | c = Logarithm of Power }} {{eqn | r = \frac 1 {n a^n} \ln \left({\frac {x^n} {x^n + a^n} }\right) + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
From Even Power is Non-Negative: :$x^{2 n} \ge 0$ The result follows from the definition of absolute value. {{qed}} Category:Absolute Value Function 7z20ha641tr6i2zwus61yrou1g6lsnr
{{begin-eqn}} {{eqn | l = u       | r = x^n + a^n       | c =  }} {{eqn | ll = \implies       | l = \frac {\mathrm d u} {\mathrm d x}       | r = n x^{n - 1}        | c = Power Rule for Derivatives and Derivative of Constant }} {{eqn | ll = \implies       | l = \int \frac {x^{n - 1} \ \mathrm d x} {x^n + a^n}       | r = \frac 1 n \ln \left \vert{x^n + a^n}\right\vert + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}