$\int \frac{x+2}{2x^3+3x^2+3x+1}\, \mathrm{d}x$
I can get it down to this:
$\int \frac{2}{2x+1} - \frac{x}{x^2+x+1}\, \mathrm{d}x $
I can solve the first part but I don't exactly follow the method in the book.
$ = \ln \vert 2x+1 \vert - \frac{1}{2}\int \frac{\left(2x+1\right) -1}{x^2+x+1}\, \mathrm{d}x $ $= \ln \vert 2x+1 \vert - \frac{1}{2} \int \frac{\mathrm{d}\left(x^2+x+1\right)}{x^2+x+1} + \frac{1}{2}\int \dfrac{\mathrm{d}x}{\left(x+\dfrac{1}{2}\right)^2 + \frac{3}{4}} $
For the 2nd part:
I tried $ u = x^2+x+1 $ and $\mathrm{d}u = 2x+1\, \mathrm{d}x$ that leaves me with $\frac{\mathrm{d}u - 1}{2} = x\, \mathrm{d}x$ which seems wrong.
because $x^2+x+1$ doesn't factor, I don't see how partial fractions again will help.
$x = Ax+B$ isn't helpful.