<s>[INST] Evaluation of $\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx$ 
Evaluate the integral
  $$\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx$$

My Attempt:
Let $f(x) = \frac{ax+b}{(x^3+3x+1)^2}.$ Now differentiate both side with respect to $x$, and we get
$$
\begin{align}
f'(x) &= \frac{(x^3+3x+1)^2\cdot a-2\cdot (x^3+3x+1)\cdot (3x^2+3)\cdot (ax+b)}{(x^3+3x+1)^4}\\
&= \frac{5x^3+3x-1}{(x^3+3x+1)^3}\cdot\frac{(x^3+3x+1)\cdot a-6\cdot (x^2+1)\cdot (ax+b)}{(x^3+3x+1)^3}\\
&= \frac{5x^3+3x-1}{(x^3+3x+1)^3} \frac{-5ax^3+6bx^2-3ax+(a-6b)}{(x^3+3x+1)^3}\\
&= \frac{5x^3+3x-1}{(x^3+3x+1)^3}
\end{align}
$$
for $a = -1$ and $b = 0$. Thus, by the Fundamental Theorem of Calculus,
$$
\begin{align}
\int\frac{5x^3+3x-1}{(x^3+3x+1)^3}\,dx &= \int f'(x)\,dx\\
&= f(x)\\
&= -\frac{x}{(x^3+3x+1)^2}+\mathcal{C}
\end{align}
$$
How we can solve the above integral directly (maybe by using the substitution method)?
 [/INST] Here is another solution... I also happen to think that it cannot be done with substitution.
$$\begin{aligned} 
\int \frac{5x^3+3x-1}{\left ( x^3+3x+1 \right )^3}\,dx &=\int \frac{-x^3-3x-1+6x^3+6x}{\left ( x^3+3x+1 \right )^3}\,dx \\  
 &= \int \frac{-x^3-3x-1+2x\left ( 3x^2+3 \right )}{\left ( x^3+x+1 \right )^3}\,dx\\  
 &= \int \frac{-\left ( x^3+3x+1 \right )^2+2x\left ( x^3+3x+1 \right )\left ( 3x^2+3x \right )}{\left ( x^3+3x+1 \right )^4}\,dx\\  
 &= \int \frac{-(x)'\left ( x^3+3x+1 \right )^2+x\left[ \left ( x^3+3x+1 \right )^2 \right]'}{\left ( x^3+3x+1 \right )^4}\,dx\\  
 &= \int \left [ -\frac{x}{\left ( x^3+3x+1 \right )^2} \right ]' \,dx = -\frac{x}{\left ( x^3+3x+1 \right )^2}+c, \; \; c \in \mathbb{R} 
\end{aligned}$$
</s>