Primitive of x squared over a x + b/Proof 2

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Theorem

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


Proof

From Reduction Formula for Primitive of $x^m \paren {a x + b}^n$: Decrement of Power of $x$:

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

Let $m = 2$ and $n = -1$.


Then:

\(\ds \int \frac {x^2 \rd x} {a x + b}\) \(=\) \(\ds \int x^2 \paren {a x + b}^{-1} \rd x\)
\(\ds \) \(=\) \(\ds \frac {x^2 \paren {a x + b}^0} {\paren 2 a} - \frac {2 b} {\paren 2 a} \int x^1 \paren {a x + b}^{-1} \rd x\)
\(\ds \) \(=\) \(\ds \frac {x^2} {2 a} - \frac b a \int \frac {x \rd x} {a x + b}\) simplifying
\(\ds \) \(=\) \(\ds \frac {x^2} {2 a} - \frac b a \paren {\frac x a - \frac b {a^2} \ln \size {a x + b} } + C\) Primitive of $x$ over $a x + b$
\(\ds \) \(=\) \(\ds \frac {x^2} {2 a} - \frac {b x} {a^2} + \frac {b^2} {a^3} \ln \size {a x + b} + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {a^2 x^2} {2 a^3} - \frac {2 a b x} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C\)
\(\ds \) \(=\) \(\ds \frac {a^2 x^2 + 2 a b x} {2 a^3} - \frac {4 a b x} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C\)
\(\ds \) \(=\) \(\ds \frac {a^2 x^2 + 2 a b x + b^2} {2 a^3} - \frac {4 a b x + b^2} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C\)
\(\ds \) \(=\) \(\ds \frac {a^2 x^2 + 2 a b x + b^2} {2 a^3} - \frac {4 a b x + 4 b^2} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C\) subsuminging $\dfrac {3 b^2} {2 a^3}$ into the arbitrary constant
\(\ds \) \(=\) \(\ds \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 a b x + 2 b^2} {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C\)
\(\ds \) \(=\) \(\ds \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 b \paren {a x + b} } {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C\)

$\blacksquare$