Primitive of x cubed over a x + b cubed

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Theorem

$\ds \int \frac {x^3 \rd x} {\paren {a x + b}^3} = \frac x {a^3} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C$


Proof

Put $u = a x + b$.

Then:

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


Then:

\(\ds \int \frac {x^3 \rd x} {\paren {a x + b}^3}\) \(=\) \(\ds \int \frac 1 a \paren {\frac {u - b} a}^3 \frac 1 {u^3} \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int \frac 1 {a^4} \paren {1 - \frac {3 b} u + \frac {3 b^2} {u^2} - \frac {b^3} {u^3} } \rd u\) multiplying out and simplification
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \int \d u - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \int \frac {\d u} {u^2} - \frac {b^3} {a^4} \int \frac {\d u} {u^3}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac u {a^4} - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \int \frac {\d u} {u^2} - \frac {b^3} {a^4} \int \frac {\d u} {u^3} + C\) Primitive of Constant
\(\ds \) \(=\) \(\ds \frac u {a^4} - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \frac {-1} u - \frac {b^3} {a^4} \frac {-1} {2 u^2} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac u {a^4} - \frac {3 b} {a^4} \ln \size u - \frac {3 b^2} {a^4 u} + \frac {b^3} {2 a^4 u^2} + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac {a x + b} {a^4} - \frac {3 b^2} {a^4 \paren {a x + b} )} + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C\) substituting for $u$ and rearranging
\(\ds \) \(=\) \(\ds \frac x {a^3} + \frac b {a^4} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C\) separating first term
\(\ds \) \(=\) \(\ds \frac x {a^3} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C\) subsuming $\dfrac b {a^4}$ into the arbitrary constant $C$

$\blacksquare$


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