# Primitive of Reciprocal of x by Root of a x + b

## Theorem

For $a > 0$ and for $x \ne 0$:

$\ds \int \frac {\d x} {x \sqrt {a x + b} } = \begin {cases} \dfrac 1 {\sqrt b} \ln \size {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C & : b > 0 \\ \dfrac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C & : b < 0 \end {cases}$

where $a x + b > 0$.

## Proof

Let:

 $\ds u$ $=$ $\ds \sqrt {a x + b}$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \frac {u^2 - b} a$

Thus:

 $\ds \map F {\sqrt {a x + b} }$ $=$ $\ds \frac 1 {x \sqrt {a x + b} }$ $\ds \leadsto \ \$ $\ds \map F u$ $=$ $\ds \paren {\frac a {u^2 - b} } \frac 1 u$

Then:

 $\ds \int \frac {\d x} {x \sqrt {a x + b} }$ $=$ $\ds \frac 2 a \int u \paren {\frac a {u^2 - b} } \frac 1 u \rd u$ Primitive of Function of $\sqrt {a x + b}$ $\ds$ $=$ $\ds 2 \int \frac {\d u} {u^2 - b}$ Primitive of Constant Multiple of Function

Let $b > 0$.

Let $d = \sqrt b$.

Then:

 $\ds 2 \int \frac {\d u} {u^2 - b}$ $=$ $\ds 2 \int \frac {\d u} {u^2 - d^2}$ $\ds$ $=$ $\ds 2 \paren {\frac 1 {2 d} } \ln \size {\frac {u - d} {u + d} } + C$ Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form $\ds$ $=$ $\ds \dfrac 1 {\sqrt b} \ln \size {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C$ substituting for $u$ and $d$

$\Box$

Let $b < 0$.

Let $d = \sqrt {-b}$.

Then:

 $\ds 2 \int \frac {\d u} {u^2 - b}$ $=$ $\ds 2 \int \frac {\d u} {u^2 - \paren {-d^2} }$ $\ds$ $=$ $\ds 2 \int \frac {\d u} {u^2 + d^2}$ $\ds$ $=$ $\ds 2 \paren {\frac 1 d} \arctan {\frac u d} + C$ Primitive of Reciprocal of $x^2 + a^2$ $\ds$ $=$ $\ds \frac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C$ substituting for $u$ and $d$

$\blacksquare$