Primitive of Reciprocal of Root of x squared plus k

Theorem

Let $k \in \R$.

Then:

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

Proof

There are three cases:

$0 \le k$

$-x^2 < k < 0$

$k < -x^2$

$(1): \quad 0 \le k$

If $0 \le k$ then $k = a^2$ for some $a \in \R$.

Then Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ applies:

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

$(2): \quad -x^2 < k < 0$

If $-x^2 < k < 0$ then:

$k = -a^2$ for some $a \in \R$

and:

$x^2 - a^2 > 0$

Then Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ applies:

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

$(3): \quad k < -x^2$

If $k < -x^2$ then:

$k = -a^2$ for some $a \in \R$

and:

$a^2 - x^2 > 0$

Then Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$ applies:

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

The result holds for all three cases.

$\blacksquare$