Primitive of x over a x + b/Proof 2
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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
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
- $\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Let $m = 1$ and $n = -1$.
Then:
\(\ds \int \frac {x \rd x} {a x + b}\) | \(=\) | \(\ds \int x^1 \paren {a x + b}^{-1} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^1 \paren {a x + b}^0} {\paren 1 a} - \frac {1 b} {\paren 1 a} \int x^0 \paren {a x + b}^{-1} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac x a - \frac b a \int \frac {\d x} {a x + b}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac x a - \frac b {a^2} \ln \size {a x + b} + C\) | Primitive of Reciprocal of $a x + b$ |
$\blacksquare$