Primitive of Root of a x + b over x

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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 1

From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$:

$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$


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

$\blacksquare$


Proof 2

Let:

\(\ds v\) \(=\) \(\ds \sqrt x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 {2 \sqrt x}\) Power Rule for Derivatives
\(\ds u\) \(=\) \(\ds \frac {2 \sqrt {a x + b} } {\sqrt x}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac {\frac {\sqrt x \cdot 2 a} {2 \sqrt{a x + b} } - \frac {2 \sqrt {a x + b} } {2 \sqrt x} } x\) Quotient Rule for Derivatives etc.
\(\ds \) \(=\) \(\ds \frac {-b} {x^{3/2} \sqrt {a x + b} }\) simplifying


From Integration by Parts:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \, \frac {\d u} {\d x} \rd x$

from which:

\(\ds \int \frac {\sqrt {a x + b} } x \rd x\) \(=\) \(\ds \int \frac {2 \sqrt {a x + b} }{\sqrt x} \frac 1 {2 \sqrt x} \rd x\)
\(\ds \) \(=\) \(\ds \frac {2 \sqrt {a x + b} } {\sqrt x} \sqrt x - \int {\sqrt x} \frac {-b} {x^{3/2} \sqrt {a x + b} } \rd x\)
\(\ds \) \(=\) \(\ds 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }\) simplification

$\blacksquare$


Also see


Sources