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

## Theorem

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

## Proof

Let:

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

Then:

 $\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }$ $=$ $\displaystyle \int \frac {2 u} {p \paren {\paren {\dfrac {u^2 - b} a} + q} a u} \rd u$ Primitive of Function of $\sqrt{a x + b}$ $\displaystyle$ $=$ $\displaystyle \int \frac {2 a u} {\paren {p \paren {u^2 - b} + a q} a u} \rd u$ simplifying $\displaystyle$ $=$ $\displaystyle 2 \int \frac {\d u} {p u^2 - b p + a q}$ simplifying $\text {(1)}: \quad$ $\displaystyle$ $=$ $\displaystyle 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}$ putting denominator into the form $z^2 - a^2$

Now let $z = \sqrt p u$

 $\displaystyle z$ $=$ $\displaystyle \sqrt p u$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d z} {\d u}$ $=$ $\displaystyle \sqrt p$

Thus:

 $\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }$ $=$ $\displaystyle 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}$ from $(1)$ $\text {(2)}: \quad$ $\displaystyle$ $=$ $\displaystyle \frac 2 {\sqrt p} \int \frac {\d z} {z^2 - \paren {\sqrt {b p - a q} }^2}$ simplifying

Let $b p - a q > 0$.

Let $d = \sqrt {b p - a q}$.

Then:

 $\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }$ $=$ $\displaystyle \frac 2 {\sqrt p} \int \frac {\d z} {z^2 - d^2}$ from $(2)$ $\displaystyle$ $=$ $\displaystyle \frac 2 {\sqrt p} \frac 1 {2 d} \ln \size {\frac {z - d} {z + d} } + C$ Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt p d} \ln \size {\frac {\sqrt p u - d} {\sqrt p u + d} } + C$ substituting for $z$ $\displaystyle$ $=$ $\displaystyle \frac 1 {\sqrt p \sqrt {b p - a q} } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$ substituting for $u$ and $d$

$\Box$

Let $b p - a q < 0$.

Let $d = \sqrt {-\paren {b p - a q} } = \sqrt {a q - b p}$.

Then:

 $\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }$ $=$ $\displaystyle \frac 2 {\sqrt p} \int \frac {\d z} {z^2 + d^2}$ from $(2)$ $\displaystyle$ $=$ $\displaystyle \frac 2 {\sqrt p} \frac 1 d \arctan {\frac z d} + C$ Primitive of $\dfrac 1 {x^2 + a^2}$ $\displaystyle$ $=$ $\displaystyle \frac 2 {\sqrt p d} \arctan {\frac {\sqrt p u} d} + C$ substituting for $z$ $\displaystyle$ $=$ $\displaystyle \frac 2 {\sqrt {a q - b p} \sqrt p} \arctan {\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C$ substituting for $u$ and $d$

$\blacksquare$