Primitive of x over a x + b cubed

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Theorem

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


Proof 1

Put $u = a x + b$.

Then:

\(\ds x\) \(=\) \(\ds \frac {u - b} a\)
\(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac 1 a\)


Then:

\(\ds \int \frac {x \rd x} {\paren {a x + b}^3}\) \(=\) \(\ds \int \frac 1 a \frac {u - b} {a u^3} \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \int \frac {\d u} {u^2} - \frac b {a^2} \int \frac {\d u} {u^3}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \frac {-1} u - \frac b {a^2} \frac {-1} {2 u^2} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {-1} {a^2 \paren {a x + b} } + \frac b {2 a^2 \paren {a x + b}^2} + C\) substituting for $u$ and rearranging

$\blacksquare$


Proof 2

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

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

where $n \ne - 1$ and $n \ne - 2$.

The result follows by setting $n = -3$.

$\blacksquare$


Sources

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