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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.86$