Primitive of Reciprocal of x by x squared plus a squared/Proof 3

Theorem

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

Proof

From Primitive of Reciprocal of x by Power of x plus Power of a:

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

So:

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

\(=\)

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

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

\(\ds \)

\(=\)

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

Absolute Value of Even Power

$\blacksquare$