<s>[INST] Solving integral without partial fraction I would like to expand my pool of integral solving skills and thus try to solve older problems again, however with a different method I had used back then, when I encountered them first. For this problem I omitted the lower and upper bound and just compute the indefinite integral (values are not important to me)
One of them was: 
\begin{align}
& \int \frac{x^2-x+2}{x^3-x^2+x-1}dx = \int \frac{x(x-1)+2}{x^2(x-1)+(x-1)}dx \\[10pt]
= {} & \int \frac{x(x-1)}{x^2(x-1)+(x-1)} dx + \int \frac{2}{x^2(x-1)+(x-1)}dx \\[10pt]
= {} & \int \frac{x}{x^2+1}dx + \int \frac{2}{(x^2+1)(x-1)} dx \\[10pt]
= {} & \frac{1}{2} \log(1+x^2) +  \int \frac{2(x+1)}{(x^2+1)(x^2-1)} dx \\[10pt]
= {} & \frac{1}{2} \log(1+x^2) + 2 \int \frac{1}{x^4-1} dx + 2 \int \frac{x}{(x^2+1)(x^2-1)} dx \\[10pt]
= {} & \frac{1}{4} (\log(1-x^2)-\log(1+x^2)) + \frac{1}{2} \log(1+x^2) + 2 \int \frac{1}{x^4-1} dx \end{align}
However now I am stuck. I am sure, if I looked up in a formula table I would find something to the last integral, but not being able to solve this, means that I am missing some kind of insight or tool to further proceed. Does someone have an idea what to do here, for the last integral?
Any constructive comment, answer is appreciated. As always thanks in advance.  
 [/INST] Be smart, add zeros!
$$I=\int dx\frac{1}{x^4-1}=\\
\int dx\frac{1+\overbrace{x^2-x^2}^{=0}}{(x^2-1)(x^2+1)}=\\\int dx\frac{1}{x^2-1}-\int dx\frac{x^2}{(x^2-1)(x^2+1)}=\\
\int dx\frac{1}{x^2-1}-\int dx\frac{x^2\overbrace{-1+1}^{=0}}{(x^2-1)(x^2+1)}=\\
\int dx\frac{1}{x^2-1}-\int dx\frac{1}{x^2+1}-\underbrace{\int dx\frac{1}{(x^2-1)(x^2+1)}}_{I}
$$
therefor
$$
2I=\int dx\frac{1}{x^2-1}-\int dx\frac{1}{x^2+1}=\text{arctanh}(x)-\arctan(x)
$$
</s>