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

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Theorem

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


Proof

Let:

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

Then:

\(\displaystyle \int \frac {p x + q} {\sqrt {a x + b} } \rd x\) \(=\) \(\displaystyle \int \paren {p \paren {\frac {u^2 - b} a} + q} \frac {2 u} {a u} \rd x\) Primitive of Function of Root of $a x + b$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a^2} \int \paren {p u^2 - b p + a q} \rd x\) Linear Combination of Integrals
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a^2} \paren {\frac {p u^3} 3 - b p z + a q z} + C\) Primitive of Power
Note that at this point we can assume $u > 0$, otherwise $\sqrt {a x + b}$ would not have been defined.
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a^2} \paren {\frac {p \paren {\sqrt {a x + b} }^3} 3 - b p \sqrt {a x + b} + a q \sqrt {a x + b} } + C\) substituting for $u$
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \paren {a p x - 2 b p + 3 a q} } {3 a^2} \sqrt {a x + b} + C\) simplification

$\blacksquare$


Sources