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

Theorem

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

$\ds \int \frac {\d x} {x^3 \paren {x^2 + a^2} }$

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds \frac {-1} {a^2 x^2} - \frac 1 {a^4} \int \frac {\d x} x + \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} + C$

$\ds $

$=$

$\ds \frac {-1} {a^2 x^2} - \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} + C$

$\ds $

$=$

$\ds \frac {-1} {a^2 x^2} - \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } + C$

$\ds $

$=$

$\ds -\frac 1 {a^2 x^2} - \frac 1 {2 a^4} \map \ln {x^2} + \frac 1 {2 a^4} \map \ln {x^2 + a^2} + C$

$\ds $

$=$

$\ds -\frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \map \ln {\frac {x^2 + a^2} {x^2} } + C$

$\blacksquare$

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 + a^2$: $14.131$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 17$: Tables of Special Indefinite Integrals: $(6)$ Integrals Involving $x^2 + a^2$: $17.6.7.$