Find $\int\frac{x^2-1}{x^4+x^2+1}dx$ 
Find
  $$
\int\frac{x^2-1}{x^4+x^2+1}dx
$$

$x^4+x^2+1>0$, so not possible to factor, so I guess there is no direct way or through partial fraction decomposition. 
$$
\int\frac{x^2-1}{x^4+x^2+1}dx=\int\frac{(x^2-1)^2}{(x^2)^3-1}dx\\
$$
I have no clue of how to solve it, is there any substitution that I can give to $x$ so that it becomes simple ?

At  the bottom, write $ x^2 + 1/x^2 = (x + 1/x)^2 -2$. Then assume $ x+ 1/x = t$. Then you're done.
Let me right the detailed solution. Divide the numerator and denominator of the integrand by $x^2.$ Then we have $$ \frac{1-1/x^2}{(x+1/x)^2 - 1}$$. Then do the substitution business. Then just integrate the following nice cute function: $$ \frac{1}{t^2 -1}$$. After integrating, we have the following function $$ 1/2 ln|\frac{t-1}{t+1}|$$. QED
