# Primitive of x over a x + b

## Theorem

$\displaystyle \int \frac {x \rd x} {a x + b} = \frac x a - \frac b {a^2} \ln \size {a x + b} + C$

## Proof 1

Put $u = a x + b$

Then:

 $\displaystyle x$ $=$ $\displaystyle \frac {u - b} a$ $\displaystyle \frac {\d x} {\d u}$ $=$ $\displaystyle \frac 1 a$

Then:

 $\displaystyle \int \frac {x \rd x} {a x + b}$ $=$ $\displaystyle \int \frac 1 a \frac {u - b} {a u} \rd u$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 {a^2} \int \rd u - \frac b {a^2} \int \frac {\d u} u$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac u {a^2} + C - \frac b {a^2} \int \frac {\d u} u$ Primitive of Constant $\displaystyle$ $=$ $\displaystyle \frac u {a^2} - \frac b {a^2} \ln \size u + C$ Primitive of Reciprocal $\displaystyle$ $=$ $\displaystyle \frac {a x + b} {a^2} - \frac b {a^2} \ln \size {a x + b} + C$ substituting for $u$ $\displaystyle$ $=$ $\displaystyle \frac x a + \frac b {a^2} - \frac b {a^2} \ln \size {a x + b} + C$ simplifying $\displaystyle$ $=$ $\displaystyle \frac x a - \frac b {a^2} \ln \size {a x + b} + C$ subsuming $\dfrac b {a^2}$ into the arbitrary constant $C$

$\blacksquare$

## Proof 2

$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$

Let $m = 1$ and $n = -1$.

Then:

 $\displaystyle \int \frac {x \rd x} {a x + b}$ $=$ $\displaystyle \int x^1 \paren {a x + b}^{-1} \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {x^1 \paren {a x + b}^0} {\paren 1 a} - \frac {1 b} {\paren 1 a} \int x^0 \paren {a x + b}^{-1} \rd x$ $\displaystyle$ $=$ $\displaystyle \frac x a - \frac b a \int \frac {\d x} {a x + b}$ simplifying $\displaystyle$ $=$ $\displaystyle \frac x a - \frac b {a^2} \ln \size {a x + b} + C$ Primitive of Reciprocal of $a x + b$

$\blacksquare$

## Proof 3

 $\displaystyle \int \frac {x \ \mathrm d x} {a x + b}$ $=$ $\displaystyle \int \frac 1 a \frac {a x \ \mathrm d x} {a x + b}$ $\displaystyle$ $=$ $\displaystyle \int \frac 1 a \frac {\left({a x + b - b}\right) \ \mathrm d x} {a x + b}$ $\displaystyle$ $=$ $\displaystyle \frac 1 a \int \frac {\left({a x + b}\right) \ \mathrm d x} {a x + b} - \frac b a \int \frac {\mathrm d x} {a x + b}$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac 1 a \int \mathrm d x - \frac b a \int \frac {\mathrm d x} {a x + b}$ simplification $\displaystyle$ $=$ $\displaystyle \frac x a - \frac b a \int \frac {\mathrm d x} {a x + b}$ Primitive of Constant $\displaystyle$ $=$ $\displaystyle \frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert + C$ Primitive of $\dfrac 1 {a x + b}$

$\blacksquare$