Primitive of x squared over Root of a x + b

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Theorem

$\ds \int \frac {x^2 \rd x} {\sqrt {a x + b} } = \frac {2 \paren {3 a^2 x^2 - 4 a b x + 8 b^2} \sqrt {a x + b} } {15 a^3}$


Proof

Let:

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


Thus:

\(\ds \map F {\sqrt {a x + b} }\) \(=\) \(\ds \frac {x^2} {\sqrt {a x + b} }\)
\(\ds \leadsto \ \ \) \(\ds \map F u\) \(=\) \(\ds \paren {\frac {u^2 - b} a}^2 \frac 1 u\)


Then:

\(\ds \int \frac {x^2 \rd x} {\sqrt {a x + b} }\) \(=\) \(\ds \frac 2 a \int u \paren {\frac {u^2 - b} a}^2 \frac 1 u \rd u\) Primitive of Function of $\sqrt {a x + b}$
\(\ds \) \(=\) \(\ds \frac 2 {a^3} \int \paren {u^4 - 2 b u^2 + b^2} \rd u\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 2 {a^3} \paren {\frac {u^5} 5 - \frac {2 b u^3} 3 + b^2 u} + C\) Primitive of Power and Primitive of Constant
\(\ds \) \(=\) \(\ds \frac 2 {a^3} \paren {\frac {\paren {a x + b}^2} 5 - \frac {2 b \paren {a x + b} } 3 + b^2} \sqrt {a x + b} + C\) substituting for $u$ and extracting common factors
\(\ds \) \(=\) \(\ds \frac 2 {15 a^2} \paren {3 a^2 x^2 + 6 a b x + 3 b^2 - 10 a b x - 10 b^2 + 15 b^2} \sqrt {a x + b} + C\) multiplying out and combining fractions
\(\ds \) \(=\) \(\ds \frac {2 \paren {3 a^2 x^2 - 4 a b x + 8 b^2} \sqrt {a x + b} } {15 a^3} + C\) simplifying

$\blacksquare$


Sources

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