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

Jump to navigation
Jump to search

## Theorem

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

## Proof

Let:

\(\displaystyle u\) | \(=\) | \(\displaystyle \sqrt {a x + b}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle \frac {u^2 - b} a\) |

Thus:

\(\displaystyle \map F {\sqrt {a x + b} }\) | \(=\) | \(\displaystyle \frac 1 {x \sqrt {a x + b} }\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map F u\) | \(=\) | \(\displaystyle \paren {\frac a {u^2 - b} } \frac 1 u\) |

Then:

\(\displaystyle \int \frac {\d x} {x \sqrt {a x + b} }\) | \(=\) | \(\displaystyle \frac 2 a \int u \paren {\frac a {u^2 - b} } \frac 1 u \rd u\) | Primitive of Function of $\sqrt {a x + b}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \int \frac {\d u} {u^2 - b}\) | Primitive of Constant Multiple of Function |

Let $b > 0$.

Let $d = \sqrt b$.

Then:

\(\displaystyle 2 \int \frac {\d u} {u^2 - b}\) | \(=\) | \(\displaystyle 2 \int \frac {\d u} {u^2 - d^2}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \paren {\frac 1 {2 d} } \map \ln {\frac {u - d} {u + d} } + C\) | Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac 1 {\sqrt b} \map \ln {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C\) | substituting for $u$ and $d$ |

$\Box$

Let $b < 0$.

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

Then:

\(\displaystyle 2 \int \frac {\d u} {u^2 - b}\) | \(=\) | \(\displaystyle 2 \int \frac {\d u} {u^2 - \paren {-d^2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \int \frac {\d u} {u^2 + d^2}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 \paren {\frac 1 d} \arctan {\frac u d} + C\) | Primitive of Reciprocal of $x^2 + a^2$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C\) | substituting for $u$ and $d$ |

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.87$