Primitive of x cubed over a x + b

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Theorem

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


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:

\(\ds \int \frac {x^3 \rd x} {a x + b}\) \(=\) \(\ds \int \frac 1 a \paren {\frac {u - b} a}^3 \frac {\d u} u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \int \frac {\paren {u - b}^3} u \rd u\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \int \paren {u^2 - 3 b u + 3 b^2 - \frac {b^3} u} \rd u\) multiplying out and simplifying
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \paren {\int u^2 \rd u - \int 3 b u \rd u + \int 3 b^2 \rd u - \int \frac {b^3 \rd u} u}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + \int 3 b^2 \rd u - \int \frac {b^3} u \rd u} + C\) Primitive of Power (two instances)
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + b^2 u - \int \frac {b^3} u \rd u} + C\) Primitive of Constant and subsuming arbitrary constant $C$
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + 3 b^2 u - b^3 \int \frac {\d u} u} + C\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + 3 b^2 u - b^3 \ln \size u} + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac {\paren {a x + b}^3} {3 a^4} - \frac {3 b \paren {a x + b}^2} {2 a^4} - \frac {3 b^2 \paren {a x + b} } {a^4} + \frac {b^3} {a^4} \ln \size {a x + b} + C\) substituting for $u$ and simplifying

$\blacksquare$


Sources

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