I need to solve:
$$\int \frac{x+1}{x^2-2x+5} dx$$
I cann see that $D>N$ so I tried to scompose the $D$ but I get:
$$x_{1,2} = \frac{2 \pm \sqrt{4-20}}{2}$$
So $\Delta < 0$ and I tried to use A and B.
$$\int \frac{x+1}{x^2-2x+5} dx = \frac{Ax + B}{x^2-2x+5}$$
But here I stops, on my book there is a hint, $A(2x-2)+B$.
Where should I found that $2x-2$, is it the derivate of $x^2-2x+5$?
Yes, the hint in your book refers to the derivative of the denominator (see below $(*)$). For a rational function with a quadratic denominator (and numerator of degree less than 2; if not: perform long division first) with a negative discriminant, this method always work. If the discriminant is positive, go with partial fractions.
In general, if $b^2-4ac <0$, you want to do the following: because $(\color{red}{ax^2+bx+c})' = \color{blue}{2ax+b}$, you look for numbers $A$ and $B$ so you can write and split the integral as follows: \begin{array}{rl} \displaystyle \int \frac{px+q}{\color{red}{ax^2+bx+c}} \,\mbox{d}x & \displaystyle = \int \frac{A(\color{blue}{2ax+b})+B}{\color{red}{ax^2+bx+c}} \,\mbox{d}x \\[8pt] & \displaystyle = A \int\frac{\color{blue}{2ax+b}}{\color{red}{ax^2+bx+c}} \,\mbox{d}x + \int\frac{B}{ax^2+bx+c} \,\mbox{d}x \\[8pt] & \displaystyle = A \ln\left( ax^2+bx+c \right) + \color{green}{ \int\frac{B}{ax^2+bx+c} \,\mbox{d}x } \end{array} And then continue completing the square in the denominator of the green integral, working towards $\arctan$.
So in this case you're looking for $A$ and $B$ such that: $$x+1 = A(2x-2)+B$$ Clearly, $A = \tfrac{1}{2}$ and then it follows that $B=2$, so rewriting: $$\frac{x+1}{x^2-2x+5} = \frac{\tfrac{1}{2}\left( 2x-2 \right) + 2}{x^2-2x+5} = \frac{1}{2}\frac{2x-2}{x^2-2x+5}+\frac{2}{x^2-2x+5}$$gives the integrals: $$\int \frac{x+1}{x^2-2x+5} \,\mbox{d}x = \frac{1}{2} \int \frac{2x-2}{x^2-2x+5} \,\mbox{d}x + \int \frac{2}{x^2-2x+5} \,\mbox{d}x$$ The first gives you $\ln$ of the denominator, the second can be reworked into an $\arctan$: $$\int \frac{2}{x^2-2x+5} \,\mbox{d}x = \int \frac{2}{(x-1)^2+4} \,\mbox{d}x =\cdots$$
$(*)$
This is a handy step because a fraction where the numerator is the derivative of the denominator, has a very simple anti-derivative (integral): the natural logarithm of the denominator! Or, symbolically: $$\int \frac{g'(x)}{g(x)} \,\mbox{d}x = \ln g(x) + C$$You can check this by taking the derivative of $\ln g(x)$, or perform a substitution $u=g(x)$ on the initial integral.