Primitive of Reciprocal of a x + b cubed

Theorem

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

Proof 1

Let $u = a x + b$.

Then:

\(\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^3}\)

\(=\)

\(\displaystyle \frac 1 a \int \frac {\mathrm d u} {u^3}\)

Primitive of Function of $a x + b$

\(\displaystyle \)

\(=\)

\(\displaystyle \frac 1 a \frac {-1} {2 u^2} + C\)

Primitive of Power

\(\displaystyle \)

\(=\)

\(\displaystyle -\frac 1 {2 a \left({a x + b}\right)^2} + C\)

substituting for $u$

$\blacksquare$

Proof 2

From Primitive of Power of $a x + b$:

$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$

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)