Primitive of x squared by Root of a x + b

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Theorem

$\ds \int x^2 \sqrt {a x + b} \rd x = \frac {2 \paren {15 a^2 x^2 - 12 a b x + 8 b^2} } {105 a^3} \sqrt {\paren {a x + b}^3} + C$


Proof

Let:

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


Then:

\(\ds \int x \sqrt {a x + b} \rd x\) \(=\) \(\ds \frac 2 a \int \paren {\frac {u^2 - b} a}^2 u^2 \rd u\) Primitive of Function of $\sqrt {a x + b}$
\(\ds \) \(=\) \(\ds \frac 2 {a^3} \paren {\int u^6 \rd u - 2 b \int u^4 \rd u + b^2 \int u^2 \rd u}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 2 {a^3} \paren {\frac {u^7} 7 - 2 b \frac {u^5} 5 + b^2 \frac {u^3} 3} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 2 {a^3} \paren {\frac {15 u^4 - 42 b u^2 + 35 b^2} {105} } u^3 + C\) common denominator and $u^3$ as a factor
\(\ds \) \(=\) \(\ds \frac 2 {a^3} \paren {\frac {15 \paren {a x + b}^2 - 42 b \paren {a x + b} + 35 b^2} {105} } \sqrt {\paren {a x + b}^3} + C\) substituting for $u$
\(\ds \) \(=\) \(\ds \frac {2 \paren {15 a^2 x^2 + 30 a b x + 15 b^2 - 42 a b x - 42 b^2 + 35 b^2} } {105 a^3} \sqrt {\paren {a x + b}^3} + C\) multiplying out
\(\ds \) \(=\) \(\ds \frac {2 \paren {15 a^2 x^2 - 12 a b x + 8 b^2} } {105 a^3} \sqrt {\paren {a x + b}^3} + C\) simplifying

$\blacksquare$


Sources

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