From the partial-fraction decomposition in Example 6.4.5, it follows that

$\int \frac{5x^4\+37x^3\+15x^2-150x\+109}{x^5\+11x^4\+21x^3-59x^2-21 x\+49}\mathit{\ⅆ}x\= \int \frac{2}{x\+1}\mathit{\ⅆ}x\+\int \frac{3x-1}{x^2\+5x-7}\mathit{\ⅆ}x\+\int \frac{5x\+4}{{\left(x^2\+5x-7\right)}^2}\mathit{\ⅆ}x$

The first integral on the right evaluates to $2 \ln \(\left|x\+1\right|\)$. Apply the ideas of Table 6.5.1 to the second integral on the right, to obtain

$\int \frac{3x-1}{x^2\+5 x-7}\mathit{\ⅆ}x\=\frac{3}{2}\int \frac{\mathrm{du}}{u}-\frac{17}{2}\int \frac{\mathrm{dz}}{z^2- {\mathrm{\σ}}^2}$ 

where $z\=x\+5\/2$ and $\mathrm{\σ}\=\sqrt{53}\/2$. From Table 6.3.1, the substitution $z\=\mathrm{\σ} \sec \left(\mathrm{\θ}\right)$ changes the second integral on the right to

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

$\int \frac{5x\+4}{{\left(x^2\+5x-7\right)}^2}\mathit{\ⅆ}x\=\frac{5}{2}\int \frac{\mathrm{du}}{u^2}-\frac{17}{2}\int \frac{\mathrm{dz}}{{\left(z^2-{\mathrm{\σ}}^2\right)}^2}$ 

where $z\=x\+5\/2$ and $\mathrm{\σ}\=\sqrt{53}\/2$. From Table 6.3.1, the substitution $z\=\mathrm{\σ} \sec \left(\mathrm{\θ}\right)$ changes the second integral on the right to

where the integral of ${\csc }^3\left(\mathrm{\θ}\right)$ is detailed in Example 6.2.10.

Consequently, the value of the given integral is the sum of the two computed integrals, the log term, and $\frac{-5\/2}{x^2\+5 x-7}$, that is,

$-\frac{867}{106\sqrt{53}}\ln \left(\left|\frac{2x\+5-\sqrt{53}}{2x\+5\+\sqrt{53}}\right|\right)\+\frac{1}{53}\frac{17x-90}{x^2\+5x-7}\+2\ln \left(\left|x\+1\right|\right)\+\frac{3}{2}\ln \left(\left|x^2\+5x-7\right|\right)$

This expression is real for all real $x$, except for $x\=-1$, $-\left(5\+\sqrt{53}\right)\/2\doteq -6.14$, and $-\left(5-\sqrt{53}\right)\/2\doteq 1.14$, where both the integrand and the integral have vertical asymptotes.

From the partial-fraction decomposition in Example 6.4.5, it follows that

$\int \frac{5x^4\+37x^3\+15x^2-150x\+109}{x^5\+11x^4\+21x^3-59x^2-21 x\+49}\mathit{\ⅆ}x\= \int \frac{2}{x\+1}\mathit{\ⅆ}x\+\int \frac{3x-1}{x^2\+5x-7}\mathit{\ⅆ}x\+\int \frac{5x\+4}{{\left(x^2\+5x-7\right)}^2}\mathit{\ⅆ}x$

Let $Q\=x^2\+5 x-7$, so in the notation of Table 6.5.2, $a\=1$, $b\=5$, $c\=-7$, $n\=1$, and $q\=-53$. Noting formulas (3) and (5) in Table 6.5.2, write

Noting formulas (3) - (6) in Table 6.5.2, write

Combining these results leads to

$\frac{3}{2}\ln \left(Q\right)-\frac{867}{106\sqrt{53}}\ln \left(\frac{2 x\+5-\sqrt{53}}{2 x\+5\+\sqrt{53}}\right)\+\frac{17 x-90}{53 Q}\+2 \ln \left(x\+1\right)$ 

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. That is probably irrelevant, as once again Maple has returned an antiderivative that is not real for any value of real $x$ other than $x\=-1$, and $x\=\left(-5 \pm \sqrt{53}\right)\/2$ where there are vertical asymptotes.

The partial-fraction decomposition of the integrand, as computed by Maple's Context Panel system, is obtained in Table 6.5.4(a) where the underlying command is convert/parfrac.

The Partial Fractions rule in the Integration Methods tutor factors the quadratic in the denominators of the partial fractions in Table 6.5.4(a) to linear factors, making more terms, and terms of greater visual complexity because the factors are $x-\left(-5 \pm \sqrt{53}\right)\/2$. Table 6.5.4(b) summarizes the five terms that the  tutor utilizes for the partial-fraction decomposition and ensuing integration.

Unfortunately, the sum of these five terms is real only for $x\in \left(-1\,1\right)$. If the arguments of the logarithmic terms were wrapped in absolute-value bars, then the solution generated by the tutor would have a greater domain.