# Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form

## Theorem

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

for $0 < a < \size x$.

### Corollary

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

## Proof

We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:

$x > a$

or:

$x < -a$

where it is assumed that $a > 0$.

Consider the arcsecant substitution:

$u = \arcsec {\dfrac x a}$

which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.

That is:

$\size x \ge a$

Hence from Shape of Secant Function, this substitution is valid for all for all $x$ such that $\size {\dfrac x a} > 1$.

 $\ds u$ $=$ $\ds \arcsec {\frac x a}$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds a \sec u$ $\ds \leadsto \ \$ $\ds \frac {\d x} {\d u}$ $=$ $\ds a \sec u \tan u$ Derivative of Secant Function

Let $x > a$.

Then:

 $\ds \int \frac 1 {\sqrt {x^2 - a^2} } \rd x$ $=$ $\ds \int \frac {a \sec u \tan u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u$ Integration by Substitution $\ds$ $=$ $\ds \frac a a \int \frac {\sec u \tan u} {\sqrt {\sec^2 u - 1} } \rd u$ simplifying $\ds$ $=$ $\ds \int \sec u \frac {\tan u} {\sqrt {\tan^2 u} } \rd u$ Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds \int \sec u \rd u$ as $x > 1$ both $\sec u$ and $\tan u$ are greater than $0$ $\ds$ $=$ $\ds \ln \size {\sec u + \tan u} + C$ Primitive of $\sec x$: Secant plus Tangent Form $\ds$ $=$ $\ds \map \ln {\sec u + \tan u} + C$ as both $\sec u$ and $\tan u$ are greater than $0$ $\ds$ $=$ $\ds \map \ln {\frac x a + \sqrt {\frac {x^2} {a^2} - 1} } + C$ substituting back: $\sec u = \dfrac x a$ and Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } + C$ rearrangement $\ds$ $=$ $\ds \map \ln {\frac {x + \sqrt {x^2 - a^2} } a} + C$ rearrangement $\ds$ $=$ $\ds \map \ln {x + \sqrt {x^2 - a^2} - \ln a} + C$ Difference of Logarithms $\ds$ $=$ $\ds \map \ln {x + \sqrt {x^2 - a^2} } + C$ subsuming $-\ln a$ into constant $\ds$ $=$ $\ds \ln \size {x + \sqrt {x^2 - a^2} } + C$ as $x + \sqrt {x^2 - a^2} > 0$ for $x > a$

Now suppose $x < -a$.

Let $z = -x$.

Then:

$\d x = -\d z$

and we then have:

 $\ds \int \frac {\d x} {\sqrt {x^2 - a^2} }$ $=$ $\ds \int \frac {-\d z} {\sqrt {\paren {-z}^2 - a^2} }$ Integration by Substitution $\ds$ $=$ $\ds -\int \frac {\d z} {\sqrt {z^2 - a^2} }$ simplifying $\ds$ $=$ $\ds -\map \ln {z + \sqrt {z^2 - a^2} } + C$ from above $\ds$ $=$ $\ds \map \ln {z - \sqrt {z^2 - a^2} } - \map \ln {a^2} + C$ Negative of $\map \ln {z + \sqrt {z^2 - a^2} }$ $\ds$ $=$ $\ds \map \ln {z - \sqrt {z^2 - a^2} } + C$ subsuming $-\map \ln {a^2}$ into constant $\ds$ $=$ $\ds \map \ln {-x - \sqrt {\paren {-x}^2 - a^2} } + C$ substituting back for $x$ $\ds$ $=$ $\ds \map \ln {-\paren {x + \sqrt {x^2 - a^2} } } + C$ simplifying $\ds$ $=$ $\ds \ln \size {x + \sqrt {x^2 - a^2} } + C$ as $x + \sqrt {x^2 - a^2} < 0$ for $x < -a$

The result follows.

$\blacksquare$

## Also presented as

Some sources present this in the form:

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

which is the same as above, except that the constant $a$ has not been subsumed into the arbitrary constant $C$.