Integrate $\int\frac{x^3+1}{x^4+x^3+x^2+x}\,dx$ How do I integrate $\displaystyle\int\dfrac{x^3+1}{x^4+x^3+x^2+x}\,\mathrm dx$?
I tried by splitting the equation in two parts like $\dfrac{x^3}{x^4+x^3+x^2+x}$ and $\dfrac1{x^4+x^3+x^2+x}$ and then cancelling out the $x$ terms from the first part and then trying to integrate further.
But this did nothing much to it, and also the other part became more difficult to solve. 
How do I solve this integration?

We first realise that both our numerator and denominator are factorable.
We can factor the numerator and denominator as:
$$\frac{(x+1)(x^2-x+1)}{(x)(x^3+x^2+x+1)}.$$
We can now multiply both the numerator and denominator by $x-1$,
we will now get:
$$\frac{(x^2-1)(x^2-x+1)}{(x)(x^4-1)}.$$
We can also factor the denominator
$$\frac{(x^2-1)(x^2-x+1)}{(x)(x^2-1)(x^2+1)}.$$
The $x^2-1$ cancels out, leaving us with
$$\frac{x^2-x+1}{(x)(x^2+1)}.$$
We can now complete the square in the numerator,
$$\frac{(x-1)^2+x}{(x)(x^2+1)}.$$
We can now split the expression apart and evaluate each part separately
$$\frac{(x-1)^2}{(x)(x^2+1)}+\frac{x}{(x)(x^2+1)}.$$
In the second part of our expression $x$ cancels out, leaving us with
$$+\frac{1}{(x^2+1)}.$$
For the first part of our expression, we can expand the $(x-1)^2$ in the numerator
$$\frac{(x-1)^2}{(x)(x^2+1)}=\frac{x^2-2x+1}{(x)(x^2+1)}$$
We now split the sum up
$$\frac{x^2+1}{(x)(x^2+1)}-\frac{2x}{(x)(x^2+1)}$$
The first part of our expression is equal to $\dfrac{1}{x}$, the second part of our expression is just $-\dfrac{2}{x^2+1}.$
Therefore, our total expression evaluated together is
$$\frac{1}{x}-\frac{2}{x^2+1}+\frac{1}{(x^2+1)}=\frac{1}{x}-\frac{1}{(x^2+1)}.$$
These two expressions have elementary integrals.
\begin{align*}
\int \frac{1}{x}-\frac{1}{(x^2+1)}\,\mathrm dx&= 
\int \frac{1}{x}\,\mathrm dx-\int \frac{1}{(x^2+1)}\,\mathrm dx\\
&=\boxed{\ln |x| - \arctan (x) + C.}
\end{align*}
Thats it!
