int (2x-5) / (x^2+2x + 2) dx Find or evaluate the integral by completing the square

We have to evaluate the integral: \int \frac{2x-5}{x^2+2x+2}dx
We can write the integral as:
\int \frac{2x-5}{x^2+2x+2}dx=\int\frac{2x-5}{(x+1)^2+1}dx
Let x+1=t
So, dx=dt
Now we can write the integral as:
\int \frac{2x-5}{(x+1)^2+1}dx=\int \frac{2(t-1)-5}{t^2+1}dt
                       =\int \frac{2t-7}{t^2+1}dt
                        =\int \frac{2t}{t^2+1}dt-\int\frac{7}{t^2+1}dt         --------------->(1)
 
Now we will first evaluate the integral \int \frac{2t}{t^2+1}dt
        
Let t^2+1=u
So, 2tdt=du
Hence we can write,
\int \frac{2tdt}{t^2+1}=\int \frac{du}{u}
            =ln(u)
             =ln(t^2+1)             
 
Now we will evaluate the second integral : \int \frac{7}{t^2+1}dt
\int \frac{7}{t^2+1}dt=7\int \frac{1}{t^2+1}dt
               =7tan^{-1}(t)
 
Substituting both these integral results in (1) we get,
\int \frac{2x-5}{x^2+2x+2}dx=ln(t^2+1)-7tan^{-1}(t)+C   where C is a constant
                      =ln((x+1)^2+1)-7tan^{-1}(x+1)+C
                       =ln(x^2+2x+2)-7tan^{-1}(x+1)+C