Primitive of x squared over a x + b cubed

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Theorem

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


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:

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

$\blacksquare$


Sources

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