Primitive of Reciprocal of a x + b cubed/Proof 2

Theorem

$\ds \int \frac {\d x} {\paren {a x + b}^3} = -\frac 1 {2 a \paren {a x + b}^2} + C$

Proof

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 $n = -3$.

$\blacksquare$