Q:

How to compute $\int \frac{x}{(x^2-4x+8)^2} \mathrm dx$?

Can someone help me to compute:
 $$\int \frac{x}{(x^2-4x+8)^2}\mathrm dx$$
And, in general, the type:
$$\int \frac{N(x)}{(x^2+px+q)^n}\mathrm dx$$
with the order of polynomial $N(x)<n$ and $n$ natural greater than 1?

A:

$$\int \frac{x}{(x^2-4x+8)^2} dx = \int\frac{x - 2 + 2}{(x^2 - 4x + 8)^2}\,dx $$
$$= \frac 12\int \frac{2x - 4}{(x^2 - 4x + 8)^2}\,dx + \int \frac 2{((x-2)^2 + 2^2)^2}\,dx$$
For the first integral, use $u = x^2 - 4x + 8 \implies du = (2x-4)\,dx$.
For the second integral, put $2\tan \theta = (x-2)\implies 2\sec^2 \theta\,d\theta = dx$.

