Primitive of a x + b over p x + q
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Theorem
- $\ds \int \frac {a x + b} {p x + q} \rd x = \frac {a x} p + \frac {b p - a q} {p^2} \ln \size {p x + q} + C$
Proof
| \(\ds \int \frac {a x + b} {p x + q} \rd x\) | \(=\) | \(\ds a \int \frac {x \rd x} {p x + q} + b \int \frac {\d x} {p x + q}\) | Linear Combination of Primitives | |||||||||||
| \(\ds \) | \(=\) | \(\ds a \paren {\frac x p - \frac q {p^2} \ln \size {p x + q} } + b \int \frac {\d x} {p x + q} + C\) | Primitive of $\dfrac x {a x + b}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a x} p - \frac {a q} {p^2} \ln \size {p x + q} + b \paren {\frac 1 p \ln \size {p x + q} } + C\) | Primitive of $\dfrac 1 {a x + b}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a x} p - \frac {a q} {p^2} \ln \size {p x + q} + \frac b p \ln \size {p x + q} + C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a x} p + \frac {b p - a q} {p^2} \ln \size {p x + q} + C\) | common denominator |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.111$
- 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 17$: Tables of Special Indefinite Integrals: $(3)$ Integrals Involving $a x + b$ and $p x + q$: $17.3.7.$