Primitive of Root of a x + b over x/Proof 1
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Theorem
- $\ds \int \frac {\sqrt {a x + b} } x \rd x = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt{a x + b} }$
Proof
- $\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
Putting $m = -1$ and $n = \dfrac 1 2$:
\(\ds \int \frac {\sqrt {a x + b} } x \rd x\) | \(=\) | \(\ds \int x^{-1} \paren {a x + b}^{1/2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^0 \paren {a x + b}^{1/2} } {\frac 1 2} + \frac {\frac 1 2 b} {\frac 1 2} \int x^{-1} \paren {a x + b}^{- 1/2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }\) | simplifying |
$\blacksquare$