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

## Theorem

$\ds \int \frac {\d x} {x \paren {x^2 + a^2}^2} = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C$

## Proof

Let:

 $\ds \int \frac {\d x} {x \paren {x^2 + a^2}^2}$ $=$ $\ds \int \paren {\frac 1 {a^4 x} - \frac x {a^4 \paren {x^2 + a^2} } - \frac x {a^2 \paren {x^2 + a^2}^2} } \rd x$ Partial Fraction Expansion $\ds$ $=$ $\ds \frac 1 {a^4} \int \frac {\d x} x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2}$ Linear Combination of Integrals $\ds$ $=$ $\ds \frac 1 {a^4} \ln x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C$ Primitive of Reciprocal $\ds$ $=$ $\ds \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C$ Primitive of $\dfrac x {x^2 + a^2}$ $\ds$ $=$ $\ds \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \paren {-\frac 1 {2 \paren {x^2 + a^2} } } + C$ Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$ $\ds$ $=$ $\ds \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C$ simplification

$\blacksquare$