Example $2$ Evaluate $9$x$27$x $12$x$33$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$33$x$22$x $=$ x$(2$x$23$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$].9$x$27$x $1$ x$(2$x $1)($x $2)=$ A x B $2$x $1$ C 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 $9$x$27$x $1=$ A$($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 $9$x$27$x $1=(2$A B $2$C$)$x$2$ 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$$namely$,9.$ Likewise, the coefficients of x are equal and the constant terms are equal. This gives the following system of equations for A$,$ B$,$ and C$.2$A B $2$C $=3$A $2$B $$ 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.$)9$x$27$x $12$x$33$x$22$x dx $=121$ x $12$x $11$ x $2$ dx $=$ 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$.$