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

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Theorem

$\ds \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2} = -\frac 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C$


Proof

Let:

\(\ds \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2}\) \(=\) \(\ds \int \paren {\frac 1 {a^4 x^2} - \frac 1 {a^4 \paren {x^2 + a^2} } - \frac 1 {a^2 \paren {x^2 + a^2}^2} } \rd x\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \int \frac {\d x} {x^2} - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \frac {-1} x - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds -\frac 1 {a^4 x} - \frac 1 {a^4} \paren {\frac 1 a \arctan \frac x a} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2} + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds -\frac 1 {a^4 x} - \frac 1 {a^5} \arctan \frac x a - \frac 1 {a^2} \paren {\frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a} + C\) Primitive of $\dfrac 1 {\paren {x^2 + a^2}^2}$
\(\ds \) \(=\) \(\ds -\frac 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C\) simplification

$\blacksquare$


Sources

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