Primitive of x over a x + b/Proof 1
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Theorem
- $\ds \int \frac {x \rd x} {a x + b} = \frac x a - \frac b {a^2} \ln \size {a x + b} + C$
Proof
Put $u = a x + b$
Then:
| \(\ds x\) | \(=\) | \(\ds \frac {u - b} a\) | ||||||||||||
| \(\ds \frac {\d x} {\d u}\) | \(=\) | \(\ds \frac 1 a\) |
Then:
| \(\ds \int \frac {x \rd x} {a x + b}\) | \(=\) | \(\ds \int \frac 1 a \frac {u - b} {a u} \rd u\) | Integration by Substitution | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {a^2} \int \rd u - \frac b {a^2} \int \frac {\d u} u\) | Linear Combination of Primitives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac u {a^2} + C - \frac b {a^2} \int \frac {\d u} u\) | Primitive of Constant | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac u {a^2} - \frac b {a^2} \ln \size u + C\) | Primitive of Reciprocal | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a x + b} {a^2} - \frac b {a^2} \ln \size {a x + b} + C\) | substituting for $u$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac x a + \frac b {a^2} - \frac b {a^2} \ln \size {a x + b} + C\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac x a - \frac b {a^2} \ln \size {a x + b} + C\) | subsuming $\dfrac b {a^2}$ into the arbitrary constant $C$ |
$\blacksquare$