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?

  • $\begingroup$ $\displaystyle \int\frac{x^3+1}{x(x^2+1)(x+1)}dx$. Now using partial fraction $\endgroup$
    – 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:


We can now multiply both the numerator and denominator by $x-1$, we will now get:


We can also factor the denominator


The $x^2-1$ cancels out, leaving us with


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


We now split the sum up


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


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!


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