Primitive of x over a x + b/Proof 2

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 $a x + b$

$\blacksquare$