Primitive of x over a x + b squared by p x + q/Corollary
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Theorem
- $\ds \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } + \frac x {a x + b} } + C$
Proof
| \(\ds \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} }\) | \(=\) | \(\ds \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C\) | Primitive of $\dfrac x {\paren {a x + b}^2 \paren {p x + q} }$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac {a x + b} {a \paren {a x + b} } + \frac {a x} {a \paren {a x + b} } } + C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac 1 a + \frac x {a x + b} } + C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } + \frac x {a x + b} } + \paren {C - \frac 1 {a \paren {b p - a q} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } + \frac x {a x + b} } + C\) | subsuming constant |
$\blacksquare$