Evaluate $\int \frac{1}{x^3+3x+1}dx$ Evaluate $$\int \frac{1}{x^3+3x+1}dx$$
I tried to evaluate it but I couldn't  do . 

The polynomial $x^3+3x+1$ has exactly $1$ real root (at $a\sim-0.32$, according to WA), and hence two complex-conjugate roots. So $x^3+3x+1=(x-a)(x^2+bx+c)$ where $b^2-4c<0$. So your steps would be the following:
1) Break $\frac1{x^3+3x+1}$ as 
$$\frac1{x^3+3x+1}=\left(\frac1{x-a}-\frac{x+(a+b)}{x^2+bx+c}\right)\frac1{a^2+ab+c}$$
(open parenthesis to check)
2) Integrate, remembering that 
$$\frac{1}{x^2+bx+c}=\frac{1}{\left(x+\frac b2\right)^2+c-\frac{b^2}4}$$
and that $b^2-4c<0$, hence $4c-b^2>0$.
