Primitive of Reciprocal of a x + b cubed

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 appears)

• 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 17$: Tables of Special Indefinite Integrals: $(1)$ Integrals Involving $a x + b$: $17.1.11.$

(in which the same mistake appears)