Primitive of x over x squared plus a squared squared

Theorem

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

Proof

Let:

\(\ds z\)

\(=\)

\(\ds x^2 + a^2\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d z} {\d x}\)

\(=\)

\(\ds 2 x\)

Derivative of Power

\(\ds \leadsto \ \ \)

\(\ds \int \frac {x \rd x} {\paren {x^2 + a^2}^2}\)

\(=\)

\(\ds \int \frac {\d z} {2 z^2}\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds -\frac 1 {2 z} + C\)

Primitive of Power

\(\ds \)

\(=\)

\(\ds -\frac 1 {2 \paren {x^2 + a^2} } + C\)

substituting for $z$

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 + a^2$: $14.133$