hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | r = \int \frac {\mathrm d x} {\left({\sqrt {\frac {\left({2 a x + b}\right)^2 + 4 a c - b^2} {4 a} } }\right)^3}       | c = Completing the Square }} {{eqn | r = \int \frac {8 a \sqrt a \ \mathrm d x} {\left({\sqrt {\left({2 a x + b}\right)^2 + 4 a c - b^2} }\right)^3}       | c = simplifying }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = z       | r = 2 a x + b }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \frac {8 a \sqrt a \ \mathrm d x} {\left({\sqrt {\left({2 a x + b}\right)^2 + 4 a c - b^2} }\right)^3}       | r = \int \frac {8 a \sqrt a \ \mathrm d z} {2 a \left({\sqrt {z^2 + 4 a c - b^2} }\right)^3}       | c = Integration by Substitution }} {{eqn | r = 4 \sqrt a \int \frac {\mathrm d z} {\left({\sqrt {z^2 + 4 a c - b^2} }\right)^3}       | c = Primitive of Constant Multiple of Function }} {{end-eqn}} Let $4 a c - b^2 > 0$. Then: {{begin-eqn}} {{eqn | l = 4 \sqrt a \int \frac {\mathrm d z} {\left({\sqrt {z^2 + 4 a c - b^2} }\right)^3}       | r = 4 \sqrt a \left({\frac z {\left({4 a c - b^2}\right) \sqrt {z^2 + 4 a c - b^2} } }\right) + C       | c = Primitive of $\dfrac 1 {\left({\sqrt {x^2 + a^2} }\right)^3}$ }} {{end-eqn}} Let $4 a c - b^2 < 0$. Then: {{begin-eqn}} {{eqn | l = 4 \sqrt a \int \frac {\mathrm d z} {\left({\sqrt {z^2 + 4 a c - b^2} }\right)^3}       | r = 4 \sqrt a \int \frac {\mathrm d z} {\left({\sqrt {z^2 - \left({b^2 - 4 a c}\right)} }\right)^3}       | c =  }} {{eqn | r = 4 \sqrt a \left({\frac {-z} {\left({b^2 - 4 a c}\right) \sqrt {z^2 - \left({b^2 - 4 a c}\right)} } }\right) + C       | c = Primitive of $\dfrac 1 {\left({\sqrt {x^2 - a^2} }\right)^3}$ }} {{eqn | r = 4 \sqrt a \left({\frac z {\left({4 a c - b^2}\right) \sqrt {z^2 + 4 a c - b^2} } }\right) + C       | c = simplifying }} {{end-eqn}} Thus in both cases the same result applies, and so: {{begin-eqn}} {{eqn | l = 4 \sqrt a \int \frac {\mathrm d z} {\left({\sqrt {z^2 + 4 a c - b^2} }\right)^3}       | r = \frac {4 \sqrt a \left({2 a x + b}\right)} {\left({4 a c - b^2}\right) \sqrt {\left({2 a x + b}\right)^2 + 4 a c - b^2} } + C       | c = substituting for $z$ and simplifying }} {{eqn | r = \frac {4 \sqrt a \left({2 a x + b}\right)} {\left({4 a c - b^2}\right) \sqrt {4 a \left({a x^2 + b x + c}\right)} } + C       | c = Completing the Square }} {{eqn | r = \frac {2 \left({2 a x + b}\right)} {\left({4 a c - b^2}\right) \sqrt {a x^2 + b x + c} } + C       | c = simplifying }} {{end-eqn}} {{qed}}
First: {{begin-eqn}} {{eqn | l = z       | r = a x^2 + b x + c       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a x + b       | c = Derivative of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {x \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | r = \frac 1 {2 a} \int \frac {2 a x \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b - b}\right) \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | c =  }} {{eqn | r = \frac 1 {2 a} \int \frac {\left({2 a x + b}\right) \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3} - \frac b {2 a} \int \frac {\mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}        | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {2 a} \int \frac {\mathrm d z} {z^{3/2} } - \frac b {2 a} \int \frac {\mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}        | c = Integration by Substitution }} {{eqn | r = \frac 1 {2 a} \left({\frac {-2} {\sqrt z} }\right) - \frac b {2 a} \int \frac {\mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | c = Primitive of Power }} {{eqn | r = \frac 1 {a \sqrt z} - \frac b {2 a} \left({\frac {2 \left({2 a x + b}\right)} {\left({4 a c - b^2}\right) \sqrt {a x^2 + b x + c} } }\right)       | c = Primitive of $\dfrac 1 {\left({\sqrt {a x^2 + b x + c} }\right)^3}$ }} {{eqn | r = \frac 1 {a \sqrt {a x^2 + b x + c} } - \frac b {2 a} \left({\frac {2 \left({2 a x + b}\right)} {\left({4 a c - b^2}\right) \sqrt {a x^2 + b x + c} } }\right)       | c = substituting for $z$ }} {{eqn | r = \frac {2 \left({b x + 2 c}\right)} {\left({b^2 - 4 a c}\right) \sqrt {a x^2 + b x + c} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \int \frac {x^2 \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | o =        | c =  }} {{eqn | r = \int \frac {a x^2 \ \mathrm d x} {a \left({\sqrt {a x^2 + b x + c} }\right)^3}       | c = multiplying top and bottom by $a$ }} {{eqn | r = \int \frac {\left({a x^2 + b x + c - b x - c}\right) \ \mathrm d x} {a\left({\sqrt {a x^2 + b x + c} }\right)^3}       | c =  }} {{eqn | r = \frac 1 a \int \frac {\left({a x^2 + b x + c}\right) \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3} - \frac b a \int \frac {x \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3} - \frac c a \int \frac {\mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 a \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} } - \frac b a \int \frac {x \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3} - \frac c a \int \frac {\mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | c = simplifying }} {{eqn | r = \frac 1 a \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | o =        | ro= -       | r = \frac b a \left({\frac {2 \left({b x + 2 c}\right)} {\left({b^2 - 4 a c}\right) \sqrt {a x^2 + b x + c} } }\right)       | c = Primitive of $\dfrac x {\left({\sqrt {a x^2 + b x + c} }\right)^3}$ }} {{eqn | o =        | ro= -       | r = \frac c a \left({\frac {2 \left({2 a x + b}\right)} {\left({4 a c - b^2}\right) \sqrt {a x^2 + b x + c} } }\right)       | c = Primitive of $\dfrac 1 {\left({\sqrt {a x^2 + b x + c} }\right)^3}$ }} {{eqn | r = \frac {\left({2 b^2 - 4 a c}\right) x + 2 b c} {a \left({4 a c - b^2}\right) \sqrt {a x^2 + b x + c} } + \frac 1 a \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \int \frac {\mathrm d x} {x \left({\sqrt {a x^2 + b x + c} }\right)^3}       | o =        | c =  }} {{eqn | r = \int \frac {c \ \mathrm d x} {c x \left({\sqrt {a x^2 + b x + c} }\right)^3}       | c = multiplying top and bottom by $c$ }} {{eqn | r = \int \frac {\left({a x^2 + b x + c - a x^2 - b x}\right) \ \mathrm d x} {c x \left({\sqrt {a x^2 + b x + c} }\right)^3}       | c =  }} {{eqn | r = \frac 1 c \int \frac {\left({a x^2 + b x + c}\right) \ \mathrm d x} {x \left({\sqrt {a x^2 + b x + c} }\right)^3} - \frac a c \int \frac {x^2 \ \mathrm d x} {x \left({\sqrt {a x^2 + b x + c} }\right)^3} - \frac b c \int \frac {x \ \mathrm d x} {x \left({\sqrt {a x^2 + b x + c} }\right)^3}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 c \int \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {x \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3} - \frac b c \int \frac {\mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3}       | c = simplifying }} {{eqn | r = \frac 1 c \int \frac {\mathrm d x} {x \sqrt {a x^2 + b x + c} }       | c =  }} {{eqn | o =        | ro= -       | r = \frac a c \left({\frac {2 \left({b x + 2 c}\right)} {\left({b^2 - 4 a c}\right) \sqrt {a x^2 + b x + c} } }\right)       | c = Primitive of $\dfrac x {\left({\sqrt {a x^2 + b x + c} }\right)^3}$ }} {{eqn | o =        | ro= -       | r = \frac b c \left({\frac {2 \left({2 a x + b}\right)} {\left({4 a c - b^2}\right) \sqrt {a x^2 + b x + c} } }\right)       | c = Primitive of $\dfrac 1 {\left({\sqrt {a x^2 + b x + c} }\right)^3}$ }} {{eqn | r =        | c = simplifying }} {{end-eqn}} {{proof wanted|last step needs rethinking -- but I have stuff to do}}
{{begin-eqn}} {{eqn | r = \int \frac {\d x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3}       | o =        | c =  }} {{eqn | r = \int \frac {c \rd x} {c x^2 \paren {\sqrt {a x^2 + b x + c} }^3}       | c = multiplying top and bottom by $c$ }} {{eqn | r = \int \frac {\paren {a x^2 + b x + c - a x^2 - b x} \rd x} {c x^2 \paren {\sqrt {a x^2 + b x + c} }^3}       | c =  }} {{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} - \frac a c \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {x \rd x} {x^2 \paren {\sqrt {a x^2 + b x + c} }^3}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 c \int \frac {\d x} {x^2 \sqrt {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} - \frac b c \int \frac {\d x} {x \paren {\sqrt {a x^2 + b x + c} }^3}       | c = simplifying }} {{eqn | r = \frac 1 c \paren {-\frac {\sqrt {a x^2 + b x + c} } {c x} - \frac b {2 c} \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } }       | c = Primitive of $\dfrac 1 {x^2 \sqrt {a x^2 + b x + c} }$ }} {{eqn | o =        | ro= -       | r = \frac a c \paren {\frac {2 \paren {2 a x + b} } {\paren {4 a c - b^2} \sqrt {a x^2 + b x + c} } }       | c = Primitive of $\dfrac 1 {\paren {\sqrt {a x^2 + b x + c} }^3}$ }} {{eqn | o =        | ro= -       | r = \frac b c \paren {\frac 1 {c \sqrt {a x^2 + b x + c} } + \frac 1 c \int \frac {\d x} {x \sqrt {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {\sqrt {a x^2 + b x + c} }^3} }       | c = Primitive of $\dfrac 1 {x \paren {\sqrt {a x^2 + b x + c} }^3}$ }} {{eqn | r =        | c = simplifying }} {{end-eqn}} {{finish|Get it into the advertised format}}
{{begin-eqn}} {{eqn | o =        | r = \int x \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x }} {{eqn | r = \int \frac {2 a x \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x} {2 a}       | c = multiplying top and bottom by $2 a$ }} {{eqn | r = \int \frac {\left({2 a x + b - b}\right) \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x} {2 a} }} {{eqn | n = 1       | r = \frac 1 {2 a} \int \left({2 a x + b}\right) \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x - \frac b {2 a} \int \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x       | c = Linear Combination of Integrals }} {{end-eqn}} Let: {{begin-eqn}} {{eqn | l = z       | r = a x^2 + b x + c }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 2 a x + b       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \left({2 a x + b}\right) \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x       | r = \int z^{n + \frac 1 2} \ \mathrm d z       | c = Integration by Substitution }} {{eqn | r = \frac {z^{n + \frac 3 2} } {n + \frac 3 2}       | c = Primitive of Power }} {{eqn | r = \frac {2 z^{n + \frac 3 2} } {2 n + 3}       | c = simplifying }} {{eqn | n = 2       | r = \frac {2 \left({a x^2 + b x + c}\right)^{n + \frac 3 2} } {2 n + 3}       | c = substituting for $z$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn | o =        | r = \int x \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x }} {{eqn | r = \frac 1 {2 a} \int \left({2 a x + b}\right) \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x - \frac b {2 a} \int \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x       | c = from $(1)$ }} {{eqn | r = \frac 1 {2 a} \left({\frac {2 \left({a x^2 + b x + c}\right)^{n + \frac 3 2} } {2 n + 3} }\right) - \frac b {2 a} \int \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x       | c = from $(2)$ }} {{eqn | r = \frac {\left({a x^2 + b x + c}\right)^{n + \frac 3 2} } {a \left({2 n + 3}\right)} - \frac b {2 a} \int \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x \rd x} {x^3 + a^3}       | r = \int \frac {\paren {x + a - a} \rd x} {x^3 + a^3}       | c =  }} {{eqn | r = \int \frac {\paren {x + a} \rd x} {x^3 + a^3} - a \int \frac {\d x} {x^3 + a^3}       | c = Linear Combination of Integrals }} {{eqn | r = \int \frac {\paren {x + a} \rd x} {\paren {x + a} \paren {x^2 - a x + a^2} } - a \int \frac {\d x} {x^3 + a^3}       | c = Sum of Two Cubes }} {{eqn | r = \int \frac {\d x} {x^2 - a x + a^2} - a \int \frac {\d x} {x^3 + a^3}       | c = simplifying }} {{eqn | r = \frac 2 {a \sqrt 3} \arctan \paren {\frac {2 x - a} {a \sqrt 3} } - a \int \frac {\d x} {x^3 + a^3}       | c = Primitive of $\dfrac 1 {x^2 - a x + a^2}$ }} {{eqn | r = \frac 2 {a \sqrt 3} \, \map \arctan {\frac {2 x - a} {a \sqrt 3} } - a \paren {\frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }       | c = Primitive of $\dfrac 1 {x^3 + a^3}$ }} {{eqn | r = \frac 2 {a \sqrt 3} \, \map \arctan {\frac {2 x - a} {a \sqrt 3} } - \frac 1 {6 a} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } - \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}       | c = simplifying }} {{eqn | r = \frac 1 {6 a} \, \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}       | c = gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^3 + a^3}       | r = \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \frac {x - 2 a} {3 a^2 \paren {x^2 - a x + a^2} } } \rd x       | c = Partial Fraction Expansion }} {{eqn | r = \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \frac {2 x - 4 a} {6 a^2 \paren {x^2 - a x + a^2} } } \rd x       | c = multiplying top and bottom by $2$ }} {{eqn | r = \int \paren {\frac 1 {3 a^2 \paren {x + a} } - \paren {\frac {2 x - a} {6 a^2 \paren {x^2 - a x + a^2} } } - \paren {\frac {3 a} {6 a^2 \paren {x^2 - a x + a^2} } } } \rd x       | c =  }} {{eqn | r = \frac 1 {3 a^2} \int \frac {\d x} {x + a} - \frac 1 {6 a^2} \int \frac {\paren {2 x - a} \rd x} {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {3 a^2} \ln \size {x + a} - \frac 1 {6 a^2} \int \frac {\paren {2 x - a} \rd x} {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}       | c = Primitive of $\dfrac 1 {a x + b}$ }} {{eqn | r = \frac 1 {3 a^2} \ln \size {x + a} - \frac 1 {6 a^2} \ln \size {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}       | c = Primitive of Function under its Derivative }} {{eqn | r = \frac 1 {6 a^2} \ln \size {\paren {x + a}^2} - \frac 1 {6 a^2} \ln \size {x^2 - a x + a^2} + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}       | c = Logarithm of Power }} {{eqn | r = \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {2 a} \int \frac {\d x} {x^2 - a x + a^2}       | c = Difference of Logarithms }} {{eqn | r = \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {2 a} \paren {\frac 2 {a \sqrt 3} \map \arctan {\frac {2 x - a} {a \sqrt 3} } }       | c = Primitive of $\dfrac 1 {x^2 - a x + a^2}$ }} {{eqn | r = \frac 1 {6 a^2} \ln \size {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}       | c = simplifying }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = \frac {\d} {\d x} \paren {x^3 + a^3}       | r = 3 x^2       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \frac {x^2 \rd x} {x^3 + a^3}       | r = \frac 1 3 \ln \size {x^3 + a^3} + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
From Primitive of Power of x less one over Power of x plus Power of a: :$\displaystyle \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$ So: {{begin-eqn}} {{eqn | l = \int \frac {x^2 \rd x} {x^3 + a^3}       | r = \frac 1 3 \ln \size {x^3 + a^3} + C       | c = Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 3$ }} {{end-eqn}} directly. {{qed}}
From Primitive of $\dfrac 1 {x \left({x^n + a^n}\right)}$: :$\displaystyle \int \frac {\mathrm d x} {x \left({x^n + a^n}\right)} = \frac 1 {n a^n} \ln \left\vert{\frac {x^n} {x^n + a^n} }\right\vert + C$ Setting $n = 3$: {{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({x^3 + a^3}\right)}       | r = \frac 1 {3 a^3} \ln \left\vert{\frac {x^3} {x^3 + a^3} }\right\vert + C       | c =  }} {{end-eqn}} directly. {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 \paren {x^3 + a^3} }       | r = \frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} }       | c = Lemma }} {{eqn | r = \frac {-1} {a^3 x} - \frac 1 {a^3} \paren {\frac 1 {6 a} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }       | c = Primitive of $\dfrac x {\paren {x^3 + a^3} }$ }} {{eqn | r = -\frac {-1} {a^3 x} - \frac 1 {6 a^4} \map \ln {\frac {x^2 - a x + a^2} {\paren {x + a}^2} } - \frac 1 {a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = x^3       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 3 x^2       | c =  }} {{eqn | ll= \implies       | l = \int \frac {\mathrm d x} {\left({x^3 + a^3}\right)^2}       | r = \int \frac {\mathrm d z} {3 x^2 \left({z + a^3}\right)^2}       | c = Integration by Substitution }} {{eqn | r = \frac 1 3 \int \frac {\mathrm d z} {z^{2/3} \left({z + a^3}\right)^2}       | c = Linear Combination of Integrals }} {{end-eqn}} Then from Primitive of $\dfrac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}$: :$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$ Here we have $a = 1, b = 0, m = \dfrac 2 3, p = 1, q = a^3, n = 2$. So: {{begin-eqn}} {{eqn | o =        | r = \frac 1 3 \int \frac {\mathrm d z} {z^{2/3} \left({z + a^3}\right)^2}       | c =  }} {{eqn | r = \frac 1 3 \left({\frac {-1} {\left({1}\right) \left({- a^3}\right)} \left({\frac 1 {z^{-1/3} \left({z + a^3}\right)^{1} } + \left({\frac 2 3 + 2 - 2}\right) \int \frac {\mathrm d z} {\left({z}\right)^{2/3} \left({z + a^3}\right)^{1} } }\right)}\right)       | c = Primitive of $\dfrac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}$ }} {{eqn | r = \frac 1 {3 a^3 z^{-1/3} \left({z + a^3}\right)} + \frac 2 {9 a^3} \int \frac {\mathrm d z} {z^{2/3} \left({z + a^3}\right)}       | c = simplifying }} {{eqn | r = \frac x {3 a^3 \left({x^3 + a^3}\right)} + \frac 2 {9 a^3} \int \frac {3 x^2 \mathrm d x} {x^2 \left({x^3 + a^3}\right)}       | c = substituting for $z$ }} {{eqn | r = \frac x {3 a^3 \left({x^3 + a^3}\right)} + \frac 2 {3 a^3} \int \frac {\mathrm d x} {\left({x^3 + a^3}\right)}       | c = simplifying }} {{eqn | r = \frac x {3 a^3 \left({x^3 + a^3}\right)} + \frac 2 {3 a^3} \left({\frac 1 {6 a^2} \ln \left({\frac {\left({x + a}\right)^2} {x^2 - a x + a^2} }\right) + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }\right)       | c = Primitive of $\dfrac 1 {\left({x^3 + a^3}\right)}$ }} {{eqn | r = \frac x {3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {9 a^5} \ln \left({\frac {\left({x + a}\right)^2} {x^2 - a x + a^2} }\right) + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = x^3       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 3 x^2       | c =  }} {{eqn | ll= \implies       | l = \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | r = \int \frac {x \ \mathrm d z} {3 x^2 \left({z + a^3}\right)^2}       | c = Integration by Substitution }} {{eqn | r = \frac 1 3 \int \frac {\mathrm d z} {z^{1/3} \left({z + a^3}\right)^2}       | c = Linear Combination of Integrals }} {{end-eqn}} Then from Primitive of $\dfrac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}$: :$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$ Here we have $a = 1, b = 0, m = \dfrac 1 3, p = 1, q = a^3, n = 2$. So: {{begin-eqn}} {{eqn | o =        | r = \frac 1 3 \int \frac {\mathrm d z} {z^{1/3} \left({z + a^3}\right)^2}       | c =  }} {{eqn | r = \frac 1 3 \left({\frac {-1} {\left({1}\right) \left({- a^3}\right)} \left({\frac 1 {z^{-2/3} \left({z + a^3}\right)^{1} } + \left({\frac 1 3 + 2 - 2}\right) \int \frac {\mathrm d z} {\left({z}\right)^{1/3} \left({z + a^3}\right)^{1} } }\right)}\right)       | c = Primitive of $\dfrac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}$ }} {{eqn | r = \frac 1 {3 a^3 z^{-2/3} \left({z + a^3}\right)} + \frac 1 {9 a^3} \int \frac {\mathrm d z} {z^{1/3} \left({z + a^3}\right)}       | c = simplifying }} {{eqn | r = \frac {x^2} {3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {9 a^3} \int \frac {3 x^2 \mathrm d x} {x \left({x^3 + a^3}\right)}       | c = substituting for $z$ }} {{eqn | r = \frac {x^2} {3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {3 a^3} \int \frac {x \ \mathrm d x} {\left({x^3 + a^3}\right)}       | c = simplifying }} {{eqn | r = \frac {x^2} {3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {3 a^3} \left({\frac 1 {6 a} \ln \left({\frac {x^2 - a x + a^2} {\left({x + a}\right)^2} }\right) + \frac 1 {a \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }\right)       | c = Primitive of $\dfrac x {\left({x^3 + a^3}\right)}$ }} {{eqn | r = \frac {x^2} {3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {18 a^4} \ln \left({\frac {x^2 - a x + a^2} {\left({x + a}\right)^2} }\right) + \frac 1 {3 a^4 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = x^3 + a^3       | c =  }} {{eqn | ll= \implies       | l = \frac {\mathrm d z} {\mathrm d x}       | r = 3 x^2       | c = Derivative of Power }} {{eqn | ll= \implies       | l = \int \frac {x^2 \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | r = \int \frac {\mathrm d z} {3 z^2}       | c = Integration by Substitution }} {{eqn | r = \frac {-1} {3 z} + C       | c = Primitive of Power }} {{eqn | r = \frac {-1} {3 \left({x^3 + a^3}\right)} + C       | c = substituting for $z$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\mathrm d x} {x \left({x^3 + a^3}\right)^2}       | r = \int \frac {a^3 \ \mathrm d x} {a^3 x \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 \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 \left({x^3 + a^3}\right)^2} - \frac 1 {a^3} \int \frac {x^3 \ \mathrm d x} {x \left({x^3 + a^3}\right)^2}       | c = Linear Combination of Integrals }} {{eqn | r = \frac 1 {a^3} \int \frac {\mathrm d x} {x \left({x^3 + a^3}\right)} - \frac 1 {a^3} \int \frac {x^2 \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | c = simplifying }} {{eqn | r = \frac 1 {a^3} \left({\frac 1 {3 a^3} \ln \left\vert{\frac {x^3} {x^3 + a^3} }\right\vert}\right) - \frac 1 {a^3} \int \frac {x^2 \ \mathrm d x} {\left({x^3 + a^3}\right)^2}       | c = Primitive of $\dfrac 1 {x \left({x^3 + a^3}\right)}$ }} {{eqn | r = \frac 1 {a^3} \left({\frac 1 {3 a^3} \ln \left\vert{\frac {x^3} {x^3 + a^3} }\right\vert}\right) - \frac 1 {a^3} \left({\frac {-1} {3 \left({x^3 + a^3}\right)} }\right)       | c = Primitive of $\dfrac {x^2} {\left({x^3 + a^3}\right)^2}$ }} {{eqn | r = \frac 1 {3 a^3 \left({x^3 + a^3}\right)} + \frac 1 {3 a^6} \ln \left\vert{\frac {x^3} {x^3 + a^3} }\right\vert       | c = simplifying }} {{end-eqn}} {{qed}}