EXAMPLE $5$ Evaluate

$6$x$25$x $+12$x$3+4$x

$$dx

$.$

SOLUTION Since

x$3+4$x $=$ x$($x$2+4)$

can't be factored further, we write

$6$x$25$x $+12$x$($x$2+4)$

$=$

A

$+$

Bx $+$ C

$.$

Multiplying by

x$($x$2+4),$

we have

$6$x$25$x $+12=$A$($x$2+4)+$

x $=$

$+$ Cx $+4$A$.$

Equating coefficients, we obtain

A $+$ B $=6$C $=54$A $=12.$

Thus

A $=,$ B $=,$ and C $=5$

and so

$6$x$25$x $+12$x$3+4$x

$$dx

$=$

x

$+$

x $5$x$2+4$

$$dx

$.$

In order to integrate the second term we split it into two parts:

x $5$x$2+4$

$$dx

$=$

xx$2+4$

$$dx

$$

$5$x$2+4$

$$dx

$.$

We make the substitution

u $=$ x$2+4$

in the first of these integrals so that

du $=2$x dx$.$

We evaluate the second integral by means of this formula with

a $=2.$

$($Remember to use absolute values where appropriate.$)$

$6$x$25$x $+12$x$($x$2+4)$

$$dx

$=$

$3$x

$$dx

$+$

$$dx

$$

$5$x$2+4$

$$dx

$=$

$+$ K