Primitive of Root of a x + b over x/Proof 2

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

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

$\blacksquare$