Primitives of Functions involving Root of a x + b

Theorem

This page gathers together the primitives of some rational functions involving $\sqrt {a x + b}$.

Primitive of $\dfrac 1 {\sqrt {a x + b} }$

$\ds \int \frac {\d x} {\sqrt{a x + b} } = \frac {2 \sqrt {a x + b} } a + C$

where $a x + b > 0$.

Primitive of $\dfrac x {\sqrt {a x + b} }$

$\ds \int \frac {x \rd x} {\sqrt {a x + b} } = \frac {2 \paren {a x - 2 b} \sqrt {a x + b} } {3 a^2}$

Primitive of $\dfrac {x^2} {\sqrt {a x + b} }$

$\ds \int \frac {x^2 \rd x} {\sqrt {a x + b} } = \frac {2 \paren {3 a^2 x^2 - 4 a b x + 8 b^2} \sqrt {a x + b} } {15 a^3}$

Primitive of $\dfrac 1 {x \sqrt {a x + b} }$

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 $a x + b > 0$.

Primitive of $\dfrac 1 {x^2 \sqrt {a x + b} }$

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

Primitive of $\sqrt {a x + b}$

$\ds \int \sqrt {a x + b} \rd x = \frac {2 \sqrt {\paren {a x + b}^3} } {3 a}$

Primitive of $x \sqrt {a x + b}$

$\ds \int x \sqrt {a x + b} \rd x = \frac {2 \paren {3 a x - 2 b} } {15 a^2} \sqrt {\paren {a x + b}^3}$

Primitive of $x^2 \sqrt {a x + b}$

$\ds \int x^2 \sqrt {a x + b} \rd x = \frac {2 \paren {15 a^2 x^2 - 12 a b x + 8 b^2} } {105 a^3} \sqrt {\paren {a x + b}^3} + C$

Primitive of $\dfrac {\sqrt {a x + b} } x$

$\ds \int \frac {\sqrt {a x + b} } x \rd x = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt{a x + b} }$

Primitive of $\dfrac {\sqrt {a x + b} } {x^2}$

$\ds \int \frac {\sqrt {a x + b} } {x^2} \rd x = -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }$

Primitive of $\dfrac {x^m} {\sqrt {a x + b} }$

$\ds \int \frac {x^m} {\sqrt{a x + b} } \rd x = \frac {2 x^m \sqrt{a x + b} } {\paren {2 m + 1} a} - \frac {2 m b} {\paren {2 m + 1} a} \int \frac {x^{m - 1} } {\sqrt{a x + b} } \rd x$

Primitive of $\dfrac 1 {x^m \sqrt {a x + b} }$

$\ds \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$

Primitive of $x^m \sqrt {a x + b}$

$\ds \int x^m \sqrt {a x + b} \rd x = \frac {2 x^m} {\paren {2 m + 3} a} \paren {\sqrt {a x + b} }^3 - \frac {2 m b} {\paren {2 m + 3} a} \int x^{m - 1} \sqrt{a x + b} \rd x$

Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$

Formulation 1

$\ds \int \frac {\sqrt {a x + b} } {x^m} \rd x = -\frac {\sqrt {a x + b} } {\paren {m - 1} x^{m - 1} } + \frac a {2 \paren {m - 1} } \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$

Formulation 2

$\ds \int \frac {\sqrt{a x + b} } {x^m} \rd x = -\frac {\paren {\sqrt{a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 5} a} {\paren {2 m - 2} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x$

Also see

• Primitives of Functions involving $\paren {\sqrt {a x + b} }^m$