Primitive of x over Root of a x + b

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Theorem

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


Proof

Let:

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


Thus:

\(\ds \map F {\sqrt {a x + b} }\) \(=\) \(\ds \frac x {\sqrt {a x + b} }\)
\(\ds \leadsto \ \ \) \(\ds \map F u\) \(=\) \(\ds \paren {\frac {u^2 - b} a} \frac 1 u\)


Then:

\(\ds \int \frac {x \rd x} {\sqrt {a x + b} }\) \(=\) \(\ds \frac 2 a \int u \paren {\frac {u^2 - b} a} \frac 1 u \rd u\) Primitive of Function of $\sqrt {a x + b}$
\(\ds \) \(=\) \(\ds \frac 2 {a^2} \int \paren {u^2 - b} \rd u\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 2 {a^2} \paren {\frac {u^3} 3 - b u} + C\) Primitive of Power and Primitive of Constant
\(\ds \) \(=\) \(\ds \frac 2 {a^2} \paren {\frac {\paren {a x + b} \sqrt {a x + b} } 3 - b \sqrt {a x + b} } + C\) substituting for $u$
\(\ds \) \(=\) \(\ds \frac 2 {3 a^2} \paren {\paren {a x + b} - 3 b} \sqrt {a x + b} + C\) extracting common factors
\(\ds \) \(=\) \(\ds \frac {2 \paren {a x - 2 b} \sqrt {a x + b} } {3 a^2} + C\) simplifying

$\blacksquare$


Sources

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