Primitive of x by Power of Root of a x + b

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Theorem

$\ds \int x \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 4} } {a^2 \paren {m + 4} } - \frac {2 b \paren {\sqrt {a x + b} }^{m + 2} } {a^2 \paren {m + 2} } + C$


Proof

Let:

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


Then:

\(\ds \int x \paren {\sqrt {a x + b} }^m \rd x\) \(=\) \(\ds \frac 2 a \int \frac {u^2 - b} a u^{m + 1} \rd x\) Primitive of Function of $a x + b$
\(\ds \) \(=\) \(\ds \frac 2 {a^2} \int u^{m + 3} \rd x - \frac {2 b} {a^2} \int u^{m + 1} \rd x\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 2 {a^2} \frac {u^{m + 4} } {m + 4} - \frac {2 b} {a^2} \frac {u^{m + 2} } {m + 2} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {2 \paren {\sqrt {a x + b} }^{m + 4} } {a^2 \paren {m + 4} } - \frac {2 b \paren {\sqrt {a x + b} }^{m + 2} } {a^2 \paren {m + 2} } + C\) substituting for $u$

$\blacksquare$


Sources