Primitive of x over a x + b cubed/Proof 1

Theorem

$\ds \int \frac {x \rd x} {\paren {a x + b}^3} = \frac {-1} {a^2 \paren {a x + b} } + \frac b {2 a^2 \paren {a x + b}^2} + C$

Put $u = a x + b$.

Then:

$\ds x$

$=$

$\ds \frac {u - b} a$

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac 1 a$

Then:

$\ds \int \frac {x \rd x} {\paren {a x + b}^3}$

$=$

$\ds \int \frac 1 a \frac {u - b} {a u^3} \rd u$

$\ds $

$=$

$\ds \frac 1 {a^2} \int \frac {\d u} {u^2} - \frac b {a^2} \int \frac {\d u} {u^3}$

$\ds $

$=$

$\ds \frac 1 {a^2} \frac {-1} u - \frac b {a^2} \frac {-1} {2 u^2} + C$

$\ds $

$=$

$\ds \frac {-1} {a^2 \paren {a x + b} } + \frac b {2 a^2 \paren {a x + b}^2} + C$

substituting for $u$ and rearranging

$\blacksquare$