Primitive of Reciprocal of x cubed plus a cubed squared

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Theorem

$\ds \int \frac {\d x} {\paren {x^3 + a^3}^2} = \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 1 {9 a^5} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}$


Proof

\(\ds z\) \(=\) \(\ds x^3\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 3 x^2\)
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\paren {x^3 + a^3}^2}\) \(=\) \(\ds \int \frac {\d z} {3 x^2 \paren {z + a^3}^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 3 \int \frac {\d z} {z^{2/3} \paren {z + a^3}^2}\) Linear Combination of Primitives


Then from Primitive of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$:

$\ds \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^n} = \frac {-1} {\paren {n - 1} \paren {b p - a q} } \paren {\frac 1 {\paren {a x + b}^{m - 1} \paren {p x + q}^{n - 1} } + a \paren {m + n - 2} \int \frac {\d x} {\paren {a x + b}^m \paren {p x + q}^{n - 1} } }$

Here we have $a = 1, b = 0, m = \dfrac 2 3, p = 1, q = a^3, n = 2$.


So:

\(\ds \) \(\) \(\ds \frac 1 3 \int \frac {\d z} {z^{2/3} \paren {z + a^3}^2}\)
\(\ds \) \(=\) \(\ds \frac 1 3 \paren {\frac {-1} {\paren 1 \paren {-a^3} } \paren {\frac 1 {z^{-1/3} \paren {z + a^3}^1} + \paren {\frac 2 3 + 2 - 2} \int \frac {\d z} {\paren z^{2/3} \paren {z + a^3}^1} } }\) Primitive of $\dfrac 1 {\paren {a x + b}^m \paren {p x + q}^n}$
\(\ds \) \(=\) \(\ds \frac 1 {3 a^3 z^{-1/3} \paren {z + a^3} } + \frac 2 {9 a^3} \int \frac {\d z} {z^{2/3} \paren {z + a^3} }\) simplifying
\(\ds \) \(=\) \(\ds \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 2 {9 a^3} \int \frac {3 x^2 \rd x} {x^2 \paren {x^3 + a^3} }\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 2 {3 a^3} \int \frac {\d x} {\paren {x^3 + a^3} }\) simplifying
\(\ds \) \(=\) \(\ds \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 2 {3 a^3} \paren {\frac 1 {6 a^2} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 1 {a^2 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3} }\) Primitive of $\dfrac 1 {\paren {x^3 + a^3} }$
\(\ds \) \(=\) \(\ds \frac x {3 a^3 \paren {x^3 + a^3} } + \frac 1 {9 a^5} \map \ln {\frac {\paren {x + a}^2} {x^2 - a x + a^2} } + \frac 2 {3 a^5 \sqrt 3} \arctan \frac {2 x - a} {a \sqrt 3}\) simplifying

$\blacksquare$


Sources

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