Primitive of x over a squared minus x squared

Theorem

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

for $x^2 < a^2$.

Proof

Let:

\(\ds z\)

\(=\)

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

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds -2 x\)

Power Rule for Derivatives

\(\ds \leadsto \ \ \)

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

\(=\)

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

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 $z > 0$

\(\ds \)

\(=\)

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

substituting for $z$

$\blacksquare$

Also see

• Primitive of $\dfrac x {x^2 - a^2}$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.164$