Primitive of x over Power of a squared minus x squared

Theorem

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

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

\(\ds \leadsto \ \ \)

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

\(=\)

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

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \frac {-1} 2 \int z^{-n} \rd z\)

Primitive of Constant Multiple of Function

\(\ds \)

\(=\)

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

Primitive of Power

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(=\)

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

substituting for $z$

$\blacksquare$

Also see

• Primitive of $\dfrac x {\paren {x^2 - a^2}^n}$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.178$