Primitive of x cubed over a x + b squared

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Theorem

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


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:

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

$\blacksquare$


Sources

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