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

(in which a mistake apppears)