Indefinite integral of rational polynomial function

Question

How do I integrate $\int \frac {(x-3)}{(x^2-2x+4)^2} dx $ I know that I have to integrate via recursion. But doing so I get to a point where I have to find $\int \frac {x^2}{(x^2-2x+4)^2} dx $ which seems even harder than the first one.

Answer 1

Hint: Write $$\int\frac{2x-2}{2(x^2-2x+4)^2}-\frac{2}{(x^2-2x+4)^2}dx$$ then substitute $$u=x^2-2x+4$$,$$s=x-1,ds=dx$$ and then $$s=\sqrt{3}\tan(p),ds=\sqrt{3}\sec^2(p)dp$$

Answer 2

Observe that $$\dfrac{d\left(\dfrac{ax+b}{x^2-2x+4}\right)}{dx}=\dfrac{-ax^2-2bx+4a+2b}{(x^2-2x+4)^2}$$

Let $-ax^2-2bx+4a+2b=A(x^2-2x+4)+x-3$

Compare the coefficients of $x^2,x$ and the constants to find $a,b,A$

$$\implies\dfrac{x-3}{(x^2-2x+4)^2}=\dfrac{d\left(\dfrac{ax+b}{x^2-2x+4}\right)}{dx}-\dfrac A{x^2-2x+4}$$

Integrate both sides wrt $x$