Primitive of x over a x + b

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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

From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:

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

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 $\dfrac 1 {a x + b}$

$\blacksquare$


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