Primitive of x over x squared minus a squared

Theorem

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

for $x^2 > a^2$.

Proof

Let:

\(\ds z\)

\(=\)

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

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds 2 x\)

Power Rule for Derivatives

\(\ds \leadsto \ \ \)

\(\ds \int \frac {\d x} {x \paren {x^2 - a^2} }\)

\(=\)

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

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int \frac {\d z} z\)

Primitive of Constant Multiple of Function

\(\ds \)

\(=\)

\(\ds \frac 1 2 \ln z + C\)

Primitive of Reciprocal: Corollary as $z > 0$

\(\ds \)

\(=\)

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

substituting for $z$

$\blacksquare$

Also see

• Primitive of $\dfrac x {a^2 - x^2}$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.145$