$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.

Please note that you have: $\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$, which on dividing numerator and denominator by $x^2$ becomes:
$\int\frac{1+1/x^2}{(x^2+1/x^2)+3+3(x-1/x)}dx$=$\int\frac{1+1/x^2}{(x-1/x)^2+5+3(x-1/x)}dx$
Now put x-1/x =t so that (1+1/$x^2$)dx=dt and thus you get:
$\int\frac{1}{t^2+5+3t}dt$;which can be rearranged to get:$\int\frac{1}{t^2+5+3t}dt$=$\int\frac{1}{(t+3/2)^2+11/4}dt$=(2/$\sqrt11$)arctan[(t+3/2)2/$\sqrt11$]+c=(2/$\sqrt11$)arctan[(x-1/x+3/2)2/$\sqrt11$]+c;where c is integration constant.
PS:x>$0$ or x<$0$,only for these x , above solution is valid.At,x=$0$,arctan[(x-1/x+3/2)2/$\sqrt11$]becomes discontinuous so Fundamental theorem doesn't hold.
