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

## 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

$\blacksquare$