Primitive of Reciprocal of a x + b cubed
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Theorem
- $\ds \int \frac {\d x} {\paren {a x + b}^3} = -\frac 1 {2 a \paren {a x + b}^2} + C$
Proof 1
Let $u = a x + b$.
Then:
\(\ds \int \frac {\rd x} {\paren {a x + b}^3}\) | \(=\) | \(\ds \frac 1 a \int \frac {\rd u} {u^3}\) | Primitive of Function of $a x + b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \frac {-1} {2 u^2} + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {2 a \paren {a x + b}^2} + C\) | substituting for $u$ |
$\blacksquare$
Proof 2
From Primitive of Power of $a x + b$:
- $\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$
where $n \ne 1$.
The result follows by setting $n = -3$.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x + b$: $14.73$
- (in which a mistake apppears)