How do I evaluate this integral ?

Question

How do you evaluate this integral $\int \frac{x+1}{(x^2-x+1)^3}dx$ ?

I found out:

$\int \frac{x+1}{(x^2-x+1)^3}dx = \frac{1}{2}\int \frac{2x-1}{(x^2-x+1)^3}dx + \int \frac{3}{(x^2-x+1)^3}dx $

First integral is nicely solvable now.

I have rearrange the second one:

$ \int \frac{3}{(x^2-x+1)^3}dx = \int \frac{3}{((x-\frac{1}{2})^2+\frac{3}{4})^3}dx=\frac{1}{64}\int \frac{3}{((2x-1)^2+3)^3}dx $

Let $y=2x-1$, then I have:

$\frac{1}{128}\int \frac{3}{(y^2+3)^3}dy$

What is this integral equeal to ?

Answer 1

At last I have solved it myself.

$ arctg(y)=\int \frac{1}{y^2+1}dy = $

Using per parters

$ =\frac{1}{y^2+1}dy-\int\frac{-2y^2}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{y^2+1-1}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{1}{y^2+1}dy-2\int\frac{1}{(y^2+1)2}dy $

Now I know what is the $\int\frac{1}{(y^2+1)^2}$ equal to.

By proceeding same steps as I did above I will get what I need.

Answer 2

You should try following substitution and factorize by $$ 3^{-\frac{1}{3}}$$

$$ \frac{y}{\sqrt{3}}=\sinh(t)$$

At a factor , you will get

$$ \int \dfrac{1}{ch(t)^5}$$

Using the expression of $\th$ and its derivative which is $\frac{d(th)}{dt}=1+{th}^2={1/ch}^2$ and a intégration by part.

I'm on my phone and tired I carry on precisely tomorrow.

Repeating the process until getting to a known résultats with arctan or ln.

So in general use INDUCTION. For those:

$$ J_n= \int \frac{1}{(1+x^2)^n} $$ expressing with (separate and by part) $$J_{n+1}$$