Primitive of p x + q over Root of a x + b

Theorem

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

Let:

$\ds u$

$=$

$\ds \sqrt{a x + b}$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \frac {u^2 - b} a$

Then:

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds \frac 2 {a^2} \paren {\frac {p u^3} 3 - b p z + a q z} + C$

Note that at this point we can assume $u > 0$, otherwise $\sqrt {a x + b}$ would not have been defined.

$\ds $

$=$

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

substituting for $u$

$\ds $

$=$

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

simplification

$\blacksquare$

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$ and $p x + q$: $14.113$