Example $2$

Evaluate

$5$x$2+9$x $12$x$3+3$x$22$x

$$dx$.$

Solution

Since the degree of the numerator is less than the degree of the denominator, we don't need to divide. We factor the denominator as

$2$x$3+3$x$22$x $=$ x$(2$x$2+3$x $2)=$ x$(2$x $1)($x $+2).$

Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the following form $[$see this case$].$

$5$x$2+9$x $1$x$(2$x $1)($x $+2)$

$=$

Ax

$+$

B$2$x $1$

$+$

C

x$+2$

To determine the values of A$,$ B$,$ and C$,$ we multiply both sides of this equation by the least common denominator,

x$(2$x $1)($x $+2),$

obtaining

$5$x$2+9$x $1=$ A$$

$2$x$1$

$($x $+2)+$ Bx$($x $+2)+$ Cx$(2$x $1).$

Expanding the right side of the equation above and writing it in the standard form for polynomials, we get

$5$x$2+9$x $1=(2$A $+$ B $+2$C$)$x$2+$

$3$A$+2$B$$C

x $2$A$.$

The polynomials on each side of the equation above are identical, so the coefficients of corresponding terms must be equal. The coefficient of

x$2$

on the right side,

$2$A $+$ B $+2$C$,$

must equal the coefficient of

x$2$

on the left side$$namely$,5.$ Likewise, the coefficients of x are equal and the constant terms are equal. This gives the following system of equations for A$,$ B$,$ and C$.$

$2$A$+$B$+2$C$=3$A$+2$B$$C$=2$A$=1$

Solving, we get

A $=$

$12$

$,$ B $=,$ and C $=,$

and so we have the following. $($Remember to use absolute values where appropriate.$)$

$5$x$2+9$x $12$x$3+3$x$22$x

$$dx

$=$

$12$

$$

$1$x

$+$

$$

$12$x $1$

$+$

$$

$1$x $+2$

$$dx

$=$

$12$ln$|$x$|+25$ln$|2$x$1|+110$ln$|$x$+2|+$C

$+$ K

In integrating the middle term we have made the mental substitution

u $=2$x $1,$

which gives

du $=2$ dx

and

dx $=$

$12$

$$du$.$