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(a) WolframAlpha shows that $$\frac{x^3 + x + 1}{x^4 - 1} = -\frac{1}{2(x^2+1)} + \frac{1}{4(x+1)} + \frac{3}{4(x-1)}\,.$$ Then, using standard antiderivatives, $$\begin{aligned}\int \frac{x^3 + x + 1}{x^4 - 1} \, dx &= \int -\frac{1}{2(x^2+1)} + \frac{1}{4(x+1)} + \frac{3}{4(x-1)} \ dx \\ &= \int -\frac{1}{2(x^2+1)} \ dx + \int \frac{1}{4(x+1)} \ dx + int \frac{3}{4(x-1)} \ dx \\ &= -\frac{1}{2} \arctan(x) + \frac{1}{4} \ln|x+1| + \frac{3}{4} \ln|x-1| + C\end{aligned}$$ Directly asking WolframAlpha to integrate $\frac{x^3 + x + 1}{x^4 - 1}$ yields $$\int \frac{x^3 + x + 1}{x^4 - 1} \, dx = \frac{1}{4} \left( 3\log(1-x) + \log(x+1) - 2\arctan{x} \right) + C\,.$$ By rearranging, distributing the $\frac{1}{4}$, and noticing that WolframAlpha does not use absolute value with integrals involving the logarithm (as well as that $|x-1| = |1-x|$), we see that these two results are the same. (b) WolframAlpha shows that $$\frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6} = x^2 + 6x + 25 + \frac{255}{2(x-3)} - \frac{39}{x-2} + \frac{5}{2(x-1)}\,.$$ Then, using standard antiderivatives, $$\begin{aligned}\int \frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6} \, dx &= \int x^2 + 6x + 25 + \frac{255}{2(x-3)} - \frac{39}{x-2} + \frac{5}{2(x-1)} \ dx \\ &= \int x^2 \ dx + \int 6x \ dx + \int 25 \ dx + \int \frac{255}{2(x-3)} \ dx - \int \frac{39}{x-2} \ dx + \int \frac{5}{2(x-1)} \ dx \\ &= \frac{x^3}{3} + 3x^2 + 25x + \frac{255}{2} \ln|x-3| - 39 \ln|x-2| + \frac{5}{2} \ln|x-1| + C\end{aligned}\,.$$ Directly asking WolframAlpha to integrate $\frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6}$ yields the same result with the caveats mentioned in part (a). (c) WolframAlpha shows that $$\frac{x^2 - x - 1}{(x-3)^3} = \frac{1}{x-3} + \frac{5}{(x-3)^2} + \frac{5}{(x-3)^3}\,.$$ Then, using standard antiderivatives, $$\begin{aligned}\int \frac{x^2 - x - 1}{(x-3)^3} \, dx &= \int \frac{1}{x-3} + \frac{5}{(x-3)^2} + \frac{5}{(x-3)^3} \ dx \\ &= \int \frac{1}{x-3} \ dx + \int \frac{5}{(x-3)^2} \ dx + \int \frac{5}{(x-3)^3} \ dx \\ &= \ln|x-3| + \int 5(x-3)^{-2} \ dx + \int 5(x-3)^{-3} \ dx \\ &= \ln|x-3| - \frac{5}{x-3} - \frac{5}{2(x-3)^2} + C\end{aligned}\,.$$ Directly asking WolframAlpha to integrate $\frac{x^2 - x - 1}{(x-3)^3}$ yields $$\int \frac{x^2 - x - 1}{(x-3)^3} \, dx = \log(x-3) - \frac{5(2x-5)}{2(x-3)^2} + C\,.$$ The logarithm part is the same, and since $$-\frac{5}{x-3} - \frac{5}{2(x-3)^2} = -\frac{5[2(x-3) + 1]}{2(x-3)^2} = -\frac{5(2x-5)}{2(x-3)^2}\,,$$ we see that the rest of the result matches as well.