From the partial-fraction decomposition in Example 6.4.4, it follows that

$\int \frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19x\+15}\mathit{\ⅆ}x\= \int \frac{3x\+2}{x^2-3x\+5}\mathit{\ⅆ}x\+\int \frac{2x-1}{x^2-2 x\+3}\mathit{\ⅆ}x$

Apply the ideas of Table 6.5.1 to the first integral on the right, to obtain

$\int \frac{3x\+2}{x^2-3x\+5}\mathit{\ⅆ}x\=\frac{3}{2}\int \frac{\mathrm{du}}{u}\+\frac{13}{2}\int \frac{\mathrm{dz}}{z^2\+{\mathrm{\σ}}^2}$ 

where $z\=x-3\/2$ and $\mathrm{\σ}\=\sqrt{11}\/2$. From Table 6.3.1, the substitution $z\=\mathrm{\σ} \tan \left(\mathrm{\θ}\right)$ changes the second integral on the right to

Apply these same ideas to the second integral arising from the partial-fraction decomposition, to obtain

$\int \frac{2x-1}{x^2-2 x\+3}\mathit{\ⅆ}x\=\int \frac{\mathrm{du}}{u}\+\int \frac{\mathrm{dz}}{z^2\+{\mathrm{\σ}}^2}$ 

where $z\=x-1$ and $\mathrm{\σ}\=\sqrt{2}$. From Table 6.3.1, the substitution $z\=\mathrm{\σ} \tan \left(\mathrm{\θ}\right)$ in the second integral on the right becomes

$\int \frac{\mathrm{dz}}{z^2\+{\mathrm{\sigma }}^2}$ = $\frac{1}{\mathrm{\σ}}\int d\mathrm{\θ}$ = $\frac{1}{\sqrt{2}} \arctan \left(\mathrm{\theta }\right)\=\frac{1}{\sqrt{2}} \arctan \left(\frac{x-1}{\sqrt{2}}\right)$ 

Consequently, the value of the given integral is

$\frac{3}{2}\ln \left(x^2-3 x\+5\right)$ $\+\frac{13}{\sqrt{11}}\arctan \left(\frac{2 x-3}{\sqrt{11}}\right)$ $\+\ln \left(x^2-2 x\+3\right)$ $\+\frac{1}{\sqrt{2}} \arctan \left(\frac{x-1}{\sqrt{2}}\right)$

where the logarithms arise from the integrals of $\mathrm{du}\/u$, and absolute values are not needed because the quadratic arguments are strictly positive.

From the partial-fraction decomposition in Example 6.4.4, it follows that

$\int \frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19x\+15}\mathit{\ⅆ}x\= \int \frac{3x\+2}{x^2-3x\+5}\mathit{\ⅆ}x\+\int \frac{2x-1}{x^2-2 x\+3}\mathit{\ⅆ}x$

Let $Q_1\=x^2-3 x\+5$ and $Q_2\=x^2-2 x\+3$ so in the notation of Table 6.5.2, $a_1\=a_2\=1$, $b_1\=-3\,b_2\=-2$, $c_1\=5\,c_2\=3$, and $q_1\=11\,q_2\=8$. Noting formulas (1) and (5) in Table 6.5.2, write

Combining these results leads to

$\frac{3}{2}\ln \left(x^2-3 x\+5\right)\+\ln \left(x^2-2 x\+3\right)\+\frac{13}{\sqrt{11}}\arctan \left(\frac{2 x-3}{\sqrt{11}}\right)\+\frac{1}{\sqrt{2}}\arctan \left(\frac{x-1}{\sqrt{2}}\right)$ 

as the value of the given indefinite integral.

Note once again that Maple integrates $1\/x$ to $\ln \left(x\right)$, not $\ln \(\left|x\right|\)$, relying on a complex constant of integration to counterbalance the logarithm of a negative number.

Table 6.5.3(a) shows the result of invoking the Partial Fractions rule in the  tutor when the Sum rule is taken as an Understood Rule.

An annotated stepwise solution for the integration of the first partial fraction appears in Table 6.5.3(b). It is generated by the  tutor with the Constant, Constant Multiple, and Sum rules taken as Understood Rules. Algebra is used to split the integral to a $\mathrm{du}\/u$ term and a $\mathrm{dx}\/Q_1$ term. Rather than apply completion of the square to this latter term, Maple makes the substitution $u\=x-3\/2$, then makes the trig substitution in terms of the tangent function. Unfortunately, Maple insists on rationalizing the reciprocal of a square root so that $\sqrt{11}\/11$ is seen, instead of the more compact $1\/\sqrt{11}$.

An annotated stepwise solution for the integration of the second partial fraction appears in Table 6.5.3(c). It is generated by the  tutor with the Constant, Constant Multiple, and Sum rules taken as Understood Rules. Algebra is used to split the integral to a $\mathrm{du}\/u$ term and a $\mathrm{dx}\/Q_2$ term. Rather than apply completion of the square to this latter term, Maple makes the substitution $u\=x-1$, then makes the trig substitution in terms of the tangent function. Unfortunately, Maple insists on rationalizing the reciprocal of a square root so that $\sqrt{2}\/2$ is seen, instead of the more compact $1\/\sqrt{2}$.