Primitive of x squared over a x + b squared by p x + q

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Theorem

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


Proof

\(\ds \) \(\) \(\ds \int \frac {x^2 \rd x} {\paren {a x + b}^2 \paren {p x + q} }\)
\(\ds \) \(=\) \(\ds \int \paren {\frac {-b^2} {a \paren {b p - a q} \paren {a x + b}^2} + \frac {q^2} {\paren {b p - a q}^2 \paren {p x + q} } + \frac {b \paren {b p - 2 a q} } {a \paren {b p - a q}^2 \paren {a x + b} } } \rd x\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \frac {-b^2} {a \paren {b p - a q} } \int \frac {\d x} {\paren {a x + b}^2} + \frac {q^2} {\paren {b p - a q}^2} \int \frac {\d x} {p x + q} + \frac {b \paren {b p - 2 a q} } {a \paren {b p - a q}^2} \int \frac {\d x} {a x + b}\) Linear Combination of Primitives


We have:

\(\ds \int \frac {\d x} {\paren {a x + b}^2}\) \(=\) \(\ds \frac {-1} {a \paren {a x + b} }\) Primitive of $\dfrac 1 {\paren {a x + b}^2}$
\(\ds \int \frac {\d x} {p x + q}\) \(=\) \(\ds \frac 1 p \ln \size {p x + q}\) Primitive of $\dfrac 1 {a x + b}$
\(\ds \int \frac {\d x} {a x + b}\) \(=\) \(\ds \frac 1 a \ln \size {a x + b}\) Primitive of $\dfrac 1 {a x + b}$


Therefore:

\(\ds \) \(\) \(\ds \int \frac {x^2 \rd x} {\paren {a x + b}^2 \paren {p x + q} }\)
\(\ds \) \(=\) \(\ds \frac {-b^2} {a \paren {b p - a q} } \frac {-1} {a \paren {a x + b} } + \frac {q^2} {\paren {b p - a q}^2} \paren {\frac 1 p \ln \size {p x + q} } + \frac {b \paren {b p - 2 a q} } {a \paren {b p - a q}^2} \paren {\frac 1 a \ln \size {a x + b} } + C\) Substitution
\(\ds \) \(=\) \(\ds \frac {b^2} {\paren {b p - a q} a^2 \paren {a x + b} } + \frac 1 {\paren {b p - a q}^2} \paren {\frac {q^2} p \ln \size {p x + q} + \frac {b \paren {b p - 2 a q} } {a^2} \ln \size {a x + b} } + C\) simplifying

$\blacksquare$


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