Primitives of Functions involving Root of a x + b and p x + q

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Primitive of $p x + q$ over $\sqrt{a x + b}$

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


Primitive of Reciprocal of $\left({p x + q}\right) \sqrt{a x + b}$

Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.

Then:

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases}

\dfrac 1 {\sqrt {p \paren {b p - a q} } } \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} } } + C & : p \paren {b p - a q} > 0 \\ \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C & : p \paren {b p - a q} < 0 \\ \end {cases}$


Primitive of $\sqrt{a x + b}$ over $p x + q$

$\ds \int \frac {\sqrt{a x + b} } {p x + q} \rd x = \begin{cases}

\dfrac {2 \sqrt{a x + b} } p + \dfrac {\sqrt {b p - a q} } {p \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 x + b} } p - \dfrac {\sqrt {a q - b p} } {p \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\ \end{cases}$


Primitive of $\left({p x + q}\right)^n \sqrt{a x + b}$

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


Primitive of Reciprocal of $\left({p x + q}\right)^n \sqrt{a x + b}$

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


Primitive of $\left({p x + q}\right)^n$ over $\sqrt{a x + b}$

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


Primitive of $\sqrt{a x + b}$ over $\left({p x + q}\right)^n$

$\ds \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x = \frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$
Retrieved from "https://proofwiki.org/w/index.php?title=Primitives_of_Functions_involving_Root_of_a_x_%2B_b_and_p_x_%2B_q&oldid=189689"