Primitive of Reciprocal of x squared by x cubed plus a cubed squared

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Theorem

$\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3}^2} = \frac {-1} {a^6 x} - \frac {x^2} {3 a^6 \paren {x^3 + a^3} } - \frac 4 {3 a^6} \int \frac {x \rd x} {x^3 + a^3}$


Proof

\(\ds \int \frac {\d x} {x^2 \paren {x^3 + a^3}^2}\) \(=\) \(\ds \int \frac {a^3 \rd x} {a^3 x^2 \paren {x^3 + a^3}^2}\) multiplying top and bottom by $a^3$
\(\ds \) \(=\) \(\ds \int \frac {\paren {x^3 + a^3 - x^3} \rd x} {a^3 x^2 \paren {x^3 + a^3}^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int \frac {\paren {x^3 + a^3} \rd x} {x^2 \paren {x^3 + a^3}^2} - \frac 1 {a^3} \int \frac {x^3 \rd x} {x^2 \paren {x^3 + a^3}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \int \frac {\d x} {x^2 \paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3}^2}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {a^3} \paren {\frac {-1} {a^3 x} - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} } } - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3}^2}\) Primitive of $\dfrac 1 {x^2 \paren {x^3 + a^3} }$: Lemma
\(\ds \) \(=\) \(\ds \frac {-1} {a^6 x} - \frac 1 {a^6} \int \frac {x \rd x} {\paren {x^3 + a^3} } - \frac 1 {a^3} \int \frac {x \rd x} {\paren {x^3 + a^3}^2}\) simplifying
\(\ds \) \(=\) \(\ds \frac {-1} {a^6 x} - \frac 1 {a^6} \int \frac {x \rd x} {\paren {x^3 + a^3} } - \frac 1 {a^3} \paren {\frac {x^2} {3 a^3 \paren {x^3 + a^3} } + \frac 1 {3 a^3} \int \frac {x \rd x} {\paren {x^3 + a^3} } }\) Primitive of $\dfrac x {\paren {x^3 + a^3}^2}$: Lemma
\(\ds \) \(=\) \(\ds \frac {-1} {a^6 x} - \frac {x^2} {3 a^6 \paren {x^3 + a^3} } - \frac 4 {3 a^6} \int \frac {x \rd x} {x^3 + a^3}\) simplifying

$\blacksquare$


Sources