Primitive of x over x squared plus a squared/Proof 2

Theorem

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

Proof

From Primitive of Power of x less one over Power of x plus Power of a:

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

So:

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

\(=\)

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

Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 2$

\(\ds \)

\(=\)

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

Absolute Value of Even Power

$\blacksquare$