Integration of this function

Question

I have done the integration of functions by completing the squares. this question is also done in the similar fashion, but I am wondering. I cannot solve the following integral

$$\int \dfrac{4x + 7}{(x^2 - 2x + 3)^2}\,dx$$

Answer 1

Does $$2 \int \dfrac{2x - 2}{(x^2 - 2x + 3)^2}\,dx + \int \dfrac{11}{((x-1)^2+\sqrt{2}^2)^2}\,dx$$ look any easier?

Answer 2

In the second integral substitute $x-1=t$ to get $\int\frac{11}{(t^2+2)^2}dt$

$=\frac{11}{2}\int\frac{2+t^2-t^2}{(t^2+2)^2}dt=\frac{11}{2}\int\frac{dt}{t^2+2}-\frac{11}{4}\int\frac{t.2t}{(t^2+2)^2}dt$

$=\frac{11}{4}\int\frac{dt}{(t/2)^2+1}-\frac{11}{4}\int\frac{t}{(t^2+2)^2}d(t^2+2)$

$=\frac{11}{2}\int\frac{d(t/2)}{(t/2)^2+1}-\frac{11}{4}\int t d\left(-\frac{1}{t^2+2}\right)$

The fist integral is $tan^{-1}$ in the second apply integrating by parts.