Primitive of x by Root of a x + b

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Theorem

$\ds \int x \sqrt {a x + b} \rd x = \frac {2 \paren {3 a x - 2 b} } {15 a^2} \sqrt {\paren {a x + b}^3}$


Proof

Let:

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


Then:

\(\ds \int x \sqrt {a x + b} \rd x\) \(=\) \(\ds \frac 2 a \int \frac {u^2 - b} a u^2 \rd u\) Primitive of Function of $\sqrt {a x + b}$
\(\ds \) \(=\) \(\ds \frac 2 {a^2} \int u^4 \rd u - \frac {2 b} {a^2} \int u^2 \rd u\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 2 {a^2} \frac {u^5} 5 - \frac {2 b} {a^2} \frac {u^3} 3 + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {6 u^5 - 10 b u^3} {15 a^2} + C\) common denominator
\(\ds \) \(=\) \(\ds \frac {\paren {6 u^2 - 10 b} u^3} {15 a^2} + C\) extracting $u^3$ as a factor
\(\ds \) \(=\) \(\ds \frac {6 \paren {a x + b} - 10 b} {15 a^2} \sqrt {\paren {a x + b}^3} + C\) substituting for $u$
\(\ds \) \(=\) \(\ds \frac {2 \paren {3 a x - 2 b} } {15 a^2} \sqrt {\paren {a x + b}^3} + C\) simplifying

$\blacksquare$


Sources