# 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?

• $\displaystyle \int\frac{x^3+1}{x(x^2+1)(x+1)}dx$. Now using partial fraction
– DXT
Oct 11 '18 at 4:41

Hint: $$\dfrac{x^3+1}{x^4+x^3+x^2+x}=\dfrac{1}{x}-\dfrac{1}{x^2+1}$$

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!