Primitive of Reciprocal of a x + b by p x + q
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Theorem
- $\ds \int \frac {\d x} {\paren {a x + b} \paren {p x + q} } = \frac 1 {b p - a q} \ln \size {\frac {p x + q} {a x + b} } + C$
where $b p \ne a q$.
Proof
| \(\ds \int \frac {\d x} {\paren {a x + b} \paren {p x + q} }\) | \(=\) | \(\ds \int \paren {\frac {-a} {\paren {b p - a q} \paren {a x + b} } + \frac p {\paren {b p - a q} \paren {p x + q} } } \rd x\) | Partial Fraction Expansion | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \paren { {-a} \int \frac 1 {a x + b} \rd x + p \int \frac 1 {p x + q} \rd x}\) | Linear Combination of Primitives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \paren {\frac {-a} a \ln \size {a x + b} + \frac p p \ln \size {p x + q} } + C\) | Primitive of $\dfrac 1 {a x + b}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \paren {\ln \size {p x + q} - \ln \size {a x + b} } + C\) | simplification | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {b p - a q} \ln \size {\frac {p x + q} {a x + b} } + C\) | Difference of Logarithms |
$\blacksquare$
Sources
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.20$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x + b$ and $p x + q$: $14.105$