Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 1

Theorem

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

Proof

Let $x > a$.

Then:

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

\(=\)

\(\ds -\frac 1 a \arcoth {\frac x a} + C\)

Primitive of Reciprocal of $x^2 - a^2$ in $\arcoth$ form

\(\ds \)

\(=\)

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

$\arcoth {\dfrac x a}$ in Logarithm Form

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(=\)

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

Logarithm of Reciprocal

\(\ds \)

\(=\)

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

as $\dfrac {x - a} {x + a} > 0$ for $x > a$

Let $x < -a$.

Let $z = -x$.

Then:

$\d x = -\d z$

and we then have:

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

\(=\)

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

Integration by Substitution

\(\ds \)

\(=\)

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

simplifying

\(\ds \)

\(=\)

\(\ds -\frac 1 {2 a} \map \ln {\frac {z - a} {z + a} } + C\)

from above

\(\ds \)

\(=\)

\(\ds \frac 1 {2 a} \map \ln {\frac {z + a} {z - a} } + C\)

Logarithm of Reciprocal

\(\ds \)

\(=\)

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

substituting $-x$ back for $z$

\(\ds \)

\(=\)

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

as $x < -a$

\(\ds \)

\(=\)

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

The result follows.

$\blacksquare$