Primitive of Reciprocal of Root of a x + b

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Theorem

$\ds \int \frac {\d x} {\sqrt{a x + b} } = \frac {2 \sqrt {a x + b} } a + C$

where $a x + b > 0$.


Proof

Put $u = \sqrt{a x + b}$.

Then:

\(\ds \int \frac {\d x} {\sqrt {a x + b} }\) \(=\) \(\ds \frac 2 a \int u \frac {\d u} u\) Primitive of Function of $\sqrt {a x + b}$
\(\ds \) \(=\) \(\ds \frac 2 a \int \rd u\)
\(\ds \) \(=\) \(\ds \frac 2 a u + C\) Primitive of Constant
\(\ds \) \(=\) \(\ds \frac {2 \sqrt {a x + b} } a + C\) substituting for $u$

$\blacksquare$


Sources