Primitive of x squared over a squared minus x squared/Inverse Hyperbolic Tangent Form

Theorem

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

for $x^2 < a^2$.

Let:

$\ds \int \frac {x^2 \rd x} {a^2 - x^2}$

$=$

$\ds \int \frac {x^2 - a^2 + a^2} {a^2 - x^2} \rd x$

$\ds $

$=$

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

$\ds $

$=$

$\ds -\int \rd x + a^2 \int \frac {\rd x} {a^2 - x^2}$

$\ds $

$=$

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

$\ds $

$=$

$\ds -x + a^2 \paren {\frac 1 a \tanh^{-1} \frac x a} + C$

$\ds $

$=$

$\ds -x + a \tanh^{-1} \frac x a + C$

simplifying

$\blacksquare$