Primitive of x over a x + b/Proof 3
Jump to navigation
Jump to search
Theorem
- $\ds \int \frac {x \rd x} {a x + b} = \frac x a - \frac b {a^2} \ln \size {a x + b} + C$
Proof
| \(\ds \int \frac {x \rd x} {a x + b}\) | \(=\) | \(\ds \int \frac 1 a \frac {a x \rd x} {a x + b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \frac 1 a \frac {\paren {a x + b - b} \rd x} {a x + b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\paren {a x + b} \rd x} {a x + b} - \frac b a \int \frac {\d x} {a x + b}\) | Linear Combination of Primitives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \int \d x - \frac b a \int \frac {\d x} {a x + b}\) | simplification | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac x a - \frac b a \int \frac {\d x} {a x + b}\) | Primitive of Constant | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac x a - \frac b {a^2} \ln \size {a x + b} + C\) | Primitive of $\dfrac 1 {a x + b}$ |
$\blacksquare$